Download SBI Clerk Quadratic Equation Questions & Answers PDF for SBI Clerk Prelims and Mains exam. Very Important SBI Clerk Quadratic Equation questions on with solutions.<\/p>\n
Download quadratic equation questions for sbi clerk pdf<\/a><\/p>\n 490 Banking mocks for Rs. 299 – Enroll here<\/a><\/p>\n Instructions<\/b><\/p>\n In the following question, two equations are given. You have to solve both the equations & find out the relationship between the variables:<\/p>\n Question 1:\u00a0<\/b>$10x^2-27x-28=0$ a)\u00a0$x > y$<\/p>\n b)\u00a0$x \\geq y$<\/p>\n c)\u00a0$x < y$<\/p>\n d)\u00a0$x \\leq y$<\/p>\n e)\u00a0x = y or No relationship can be established<\/p>\n Question 2:\u00a0<\/b>$6x^2-5x+1=0$ a)\u00a0$x > y$<\/p>\n b)\u00a0$x \\geq y$<\/p>\n c)\u00a0$x < y$<\/p>\n d)\u00a0$x \\leq y$<\/p>\n e)\u00a0x = y or No relationship can be established<\/p>\n Instructions<\/b><\/p>\n In the following question, two equations numbered I and II are given. You have to solve both the equations & find out the relationship between the variables:<\/p>\n Question 3:\u00a0<\/b>$8x^2-10x+3=0$ a)\u00a0$x > y$<\/p>\n b)\u00a0$x \\geq y$<\/p>\n c)\u00a0$x < y$<\/p>\n d)\u00a0$x \\leq y$<\/p>\n e)\u00a0x = y or No relationship can be established<\/p>\n Download SBI Clerk Previous Papers PDF<\/a><\/p>\n Take a free mock test for SBI Clerk<\/a><\/p>\n Question 4:\u00a0<\/b>$2x^2-17x+35=0$ a)\u00a0$x > y$<\/p>\n b)\u00a0$x \\geq y$<\/p>\n c)\u00a0$x < y$<\/p>\n d)\u00a0$x \\leq y$<\/p>\n e)\u00a0x = y or No relationship can be established<\/p>\n Question 5:\u00a0<\/b>$x^2-40x+391=0$ a)\u00a0$x > y$<\/p>\n b)\u00a0$x \\geq y$<\/p>\n c)\u00a0$x < y$<\/p>\n d)\u00a0$x \\leq y$<\/p>\n e)\u00a0x = y or No relationship can be established<\/p>\n Question 6:\u00a0<\/b>$4x^2-11x-3=0$ a)\u00a0$x > y$<\/p>\n b)\u00a0$x \\geq y$<\/p>\n c)\u00a0$x < y$<\/p>\n d)\u00a0$x \\leq y$<\/p>\n e)\u00a0x = y or No relationship can be established<\/p>\n Question 7:\u00a0<\/b>$18x^2+3x-28=0$ a)\u00a0$x > y$<\/p>\n b)\u00a0$x \\geq y$<\/p>\n c)\u00a0$x < y$<\/p>\n d)\u00a0$x \\leq y$<\/p>\n e)\u00a0x = y or No relationship can be established<\/p>\n Banking Study Material – 18000 Questions<\/a><\/p>\n Daily Free SBI Online Tests<\/a><\/p>\n Instructions<\/b><\/p>\n In the following question, two equations numbered I and II are given. You have to solve both the equations & find out the relationship between the variables:<\/p>\n Question 8:\u00a0<\/b>$ 2x^2-11x+15 = 0 $ a)\u00a0$x > y$<\/p>\n b)\u00a0$x \\geq y$<\/p>\n c)\u00a0$x < y$<\/p>\n d)\u00a0$x \\leq y$<\/p>\n e)\u00a0x = y or No relationship can be established<\/p>\n Question 9:\u00a0<\/b>$ 15x^2+x-2 = 0 $ a)\u00a0$x > y$<\/p>\n b)\u00a0$x \\geq y$<\/p>\n c)\u00a0$x < y$<\/p>\n d)\u00a0$x \\leq y$<\/p>\n e)\u00a0x = y or No relationship can be established<\/p>\n Question 10:\u00a0<\/b>$ x^2-7x+12 = 0 $ a)\u00a0$x > y$<\/p>\n b)\u00a0$x \\geq y$<\/p>\n c)\u00a0$x < y$<\/p>\n d)\u00a0$x \\leq y$<\/p>\n e)\u00a0x = y or No relationship can be established<\/p>\n 100+ Free GK Tests<\/a><\/p>\n Question 11:\u00a0<\/b>$ 6x^2-11x+3 = 0 $ a)\u00a0$x > y$<\/p>\n b)\u00a0$x \\geq y$<\/p>\n c)\u00a0$x < y$<\/p>\n d)\u00a0$x \\leq y$<\/p>\n e)\u00a0x = y or No relationship can be established<\/p>\n Question 12:\u00a0<\/b>$ 15x^2-14x-8 =0 $ a)\u00a0$x > y$<\/p>\n b)\u00a0$x \\geq y$<\/p>\n c)\u00a0$x < y$<\/p>\n d)\u00a0$x \\leq y$<\/p>\n e)\u00a0x = y or No relationship can be established<\/p>\n Instructions<\/b><\/p>\n In the following question, two equations numbered I and II are given. You have to solve both the equations & find out the relationship between the variables:<\/p>\n Question 13:\u00a0<\/b>$2x^2-11x+15 = 0$ a)\u00a0$x > y$<\/p>\n b)\u00a0$x \\geq y$<\/p>\n c)\u00a0$x < y$<\/p>\n d)\u00a0$x \\leq y$<\/p>\n e)\u00a0x = y or No relationship can be established<\/p>\n Question 14:\u00a0<\/b>$6x^2-5x-4 = 0$ a)\u00a0$x > y$<\/p>\n b)\u00a0$x \\geq y$<\/p>\n c)\u00a0$x < y$<\/p>\n d)\u00a0$x \\leq y$<\/p>\n e)\u00a0x = y or No relationship can be established<\/p>\n Question 15:\u00a0<\/b>$12x^2-25x+12 = 0$ a)\u00a0$x > y$<\/p>\n b)\u00a0$x \\geq y$<\/p>\n c)\u00a0$x < y$<\/p>\n d)\u00a0$x \\leq y$<\/p>\n e)\u00a0x = y or No relationship can be established<\/p>\n Question 16:\u00a0<\/b>$12x^2-41x+35 = 0$ a)\u00a0$x > y$<\/p>\n b)\u00a0$x \\geq y$<\/p>\n c)\u00a0$x < y$<\/p>\n d)\u00a0$x \\leq y$<\/p>\n e)\u00a0x = y or No relationship can be established<\/p>\n Question 17:\u00a0<\/b>$9x^2-18x+5 = 0$ a)\u00a0$x > y$<\/p>\n b)\u00a0$x \\geq y$<\/p>\n c)\u00a0$x < y$<\/p>\n d)\u00a0$x \\leq y$<\/p>\n e)\u00a0x = y or No relationship can be established<\/p>\n Instructions<\/b><\/p>\n In the following question, two equations numbered I and II are given. You have to solve both the equations & find out the relationship between the variables:<\/p>\n Question 18:\u00a0<\/b>$12x^2-19x+5 = 0$ a)\u00a0$x > y$<\/p>\n b)\u00a0$x \\geq y$<\/p>\n c)\u00a0$x < y$<\/p>\n d)\u00a0$x \\leq y$<\/p>\n e)\u00a0x = y or No relationship can be established<\/p>\n Question 19:\u00a0<\/b>$9x^2-15x+4 = 0$ a)\u00a0$x > y$<\/p>\n b)\u00a0$x \\geq y$<\/p>\n c)\u00a0$x < y$<\/p>\n d)\u00a0$x \\leq y$<\/p>\n e)\u00a0x = y or No relationship can be established<\/p>\n Question 20:\u00a0<\/b>$5x^2-11x+2 = 0$ a)\u00a0$x > y$<\/p>\n b)\u00a0$x \\geq y$<\/p>\n c)\u00a0$x < y$<\/p>\n d)\u00a0$x \\leq y$<\/p>\n e)\u00a0x = y or No relationship can be established<\/p>\n General Knowledge Questions & Answers PDF<\/a><\/p>\n 490 Banking mocks for Rs. 299 – Enroll here<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(E)<\/strong><\/p>\n $10x^2-27x-28=0$ can be written as, 2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n $6x^2-5x+1=0$ can be written as, 3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n $8x^2-10x+3=0$ can be written as, 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $2x^2-17x+35=0$ can be written as, 5)\u00a0Answer\u00a0(E)<\/strong><\/p>\n $x^2-40x+391=0$ can be written as, 6)\u00a0Answer\u00a0(E)<\/strong><\/p>\n $4x^2-11x-3=0$ can be written as, 7)\u00a0Answer\u00a0(E)<\/strong><\/p>\n $18x^2+3x-28=0$ can be written as, 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n $ 2x^2-11x+15 = 0 $ can be written as $(x-\\frac{5}{2})(x-3)= 0 $ therefore, x = $\\frac{5}{2}$ or $ 3 $ 9)\u00a0Answer\u00a0(C)<\/strong><\/p>\n $ 15x^2+x-2 = 0 $ can be written as $(x-\\frac{1}{3})(x+\\frac{2}{5})= 0 $ therefore, x = $\\frac{1}{3}$ or $ \\frac{-2}{5} $ 10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n $ x^2-7x+12 = 0 $ can be written as $(x-4)(x-3)= 0 $ therefore, x = $ 4 $ or $ 3 $ 11)\u00a0Answer\u00a0(E)<\/strong><\/p>\n $ 6x^2-11x+3 = 0 $ can be written as $(x-\\frac{3}{2})(x-\\frac{1}{3})= 0 $ therefore, x = $\\frac{3}{2}$ or $ \\frac{1}{3} $ 12)\u00a0Answer\u00a0(E)<\/strong><\/p>\n $ 15x^2-14x-8 =0 $ can be written as $(x-\\frac{4}{3})(x+\\frac{2}{5})= 0 $ therefore, x = $\\frac{4}{3}$ or $ \\frac{-2}{5} $ 13)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $2x^2-11x+15 = 0$ can be written as $(x-\\frac{5}{2})(x-3) = 0$ i.e. x = $\\frac{5}{2}$ or $3$ 14)\u00a0Answer\u00a0(E)<\/strong><\/p>\n $6x^2-5x-4 = 0$ can be written as $(x+\\frac{1}{2})(x-\\frac{4}{3}) = 0$ i.e. x = -$\\frac{1}{2} or \\frac{4}{3}$ 15)\u00a0Answer\u00a0(E)<\/strong><\/p>\n $12x^2-25x+12 = 0$ can be written as $(x-\\frac{3}{4})(x-\\frac{4}{3}) = 0$ i.e. x = $\\frac{3}{4} or \\frac{4}{3}$ 16)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $12x^2-41x+35 = 0$ can be written as $(x-\\frac{7}{4})(x-\\frac{5}{3}) = 0$ i.e. x = $\\frac{7}{4} or \\frac{5}{3}$ 17)\u00a0Answer\u00a0(E)<\/strong><\/p>\n $9x^2-18x+5 = 0$ can be written as $(x-\\frac{5}{3})(x-\\frac{1}{3}) = 0$ i.e. x = $\\frac{5}{3} or \\frac{1}{3}$ 18)\u00a0Answer\u00a0(D)<\/strong><\/p>\n $12x^2-19x+5 = 0$ can be written as $(x-\\frac{5}{4})(x-\\frac{1}{3}) = 0$ 19)\u00a0Answer\u00a0(E)<\/strong><\/p>\n $9x^2-15x+4 = 0$ can be written as $(x-\\frac{1}{3})(x-\\frac{4}{3}) = 0$ i.e. x = $\\frac{1}{3} or \\frac{4}{3}$ 20)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $5x^2-11x+2 = 0$ can be written as $(x-\\frac{1}{5})(x-2) = 0$
\n$6y^2-17y-14=0$<\/p>\n
\n$y^2-7y+12=0$<\/p>\n
\n$6y^2-23y+20=0$<\/p>\n
\n$12y^2-11y-5=0$<\/p>\n
\n$4y^2-180y+2021=0$<\/p>\n
\n$6y^2-29y+35=0$<\/p>\n
\n$30y^2-47y+14=0$<\/p>\n
\n$ 2y^2-9y+10 = 0 $<\/p>\n
\n$ 20y^2-23y+6 = 0 $<\/p>\n
\n$ 8y^2-70y+153 = 0 $<\/p>\n
\n$ 3y^2-16y+5 = 0 $<\/p>\n
\n$ 10y^2-17y+3 = 0 $<\/p>\n
\n$2y^2-7y+6 = 0$<\/p>\n
\n$6y^2-11y+4 = 0$<\/p>\n
\n$12y^2-11y-5 = 0$<\/p>\n
\n$8y^2-14y+5 = 0$<\/p>\n
\n$12y^2-19y+5 = 0$<\/p>\n
\n$20y^2-57y+40 = 0$<\/p>\n
\n$12y^2-19y+5 = 0$<\/p>\n
\n$21y^2+4y-1 = 0$<\/p>\n
\n$(x-\\frac{7}{2})(x+\\frac{4}{5})=0$
\nSo $x = \\frac{7}{2}$ or $x = -\\frac{4}{5}$
\n$6y^2-17y-14=0$ can be written as,
\n$(y+\\frac{2}{3})(y-\\frac{7}{2})=0$
\nSo $y = -\\frac{2}{3}$ or $y = \\frac{7}{2}$
\nx = y or No relationship can be established
\nHence, option E is the correct choice.<\/p>\n
\n$(x-\\frac{1}{2})(x-\\frac{1}{3})=0$
\nSo $x = \\frac{1}{2}$ or $x = \\frac{1}{3}$
\n$y^2-7y+12=0$ can be written as,
\n$(y-3)(y-4)=0$
\nSo $y = 3$ or $y = 4$
\nc: $x < y$
\nHence, option C is the correct choice.<\/p>\n
\n$(x-\\frac{1}{2})(x-\\frac{3}{4})=0$
\nSo $x = \\frac{1}{2}$ or $x = \\frac{3}{4}$
\n$6y^2-23y+20=0$ can be written as,
\n$(y-\\frac{4}{3})(y-\\frac{5}{2})=0$
\nSo $y = \\frac{4}{3}$ or $y = \\frac{5}{2}$
\nSo $x < y$
\nHence, option C is the correct choice.<\/p>\n
\n$(x-5)(x-\\frac{7}{2})=0$
\nSo $x = 5$ or $x = \\frac{7}{2}$
\n$12y^2-11y-5=0$ can be written as,
\n$(y-\\frac{5}{4})(y+\\frac{1}{3})=0$
\nSo $y = \\frac{5}{4}$ or $y = -\\frac{1}{3}$
\nSo $x > y$
\nHence, option A is the correct choice.<\/p>\n
\n$(x-23)(x-17)=0$
\nSo $x = 23$ or $x = 17$
\n$4y^2-180y+2021=0$ can be written as,
\n$(y-\\frac{47}{2})(y-\\frac{43}{2})=0$
\nSo $y = \\frac{47}{2}$ or $y = \\frac{43}{2}$
\nSo No relation can be established.
\nHence, option E is the correct choice.<\/p>\n
\n$(x-3)(x+\\frac{1}{4})=0$
\nSo $x = 3$ or $x = -\\frac{1}{4}$
\n$6y^2-29y+35=0$ can be written as,
\n$(y-\\frac{5}{2})(y-\\frac{7}{3})=0$
\nSo $y = \\frac{5}{2}$ or $y = \\frac{7}{2}$
\nSo, No relationship can be established
\nHence, option E is the correct choice.<\/p>\n
\n$(x+\\frac{4}{3})(x-\\frac{7}{6})=0$
\nSo $x = -\\frac{4}{3}$ or $x = \\frac{7}{6}$
\n$30y^2-47y+14=0$ can be written as,
\n$(y-\\frac{7}{6})(y-\\frac{2}{5})=0$
\nSo $y = \\frac{7}{6}$ or $y = \\frac{2}{5}$
\nSo No relation can be established
\nHence, option E is the correct choice.<\/p>\n
\n$ 2y^2-9y+10 = 0 $ can be written as $ (y-\\frac{5}{2})(y-2) =0 $
\nTherefore, y = $\\frac{5}{2} $ or $ 2 $
\nTherefore, $x \\geq y$
\nHence, option B is the correct answer.<\/p>\n
\n$ 20y^2-23y+6 = 0 $ can be written as $ (y-\\frac{2}{5})(y-\\frac{3}{4}) =0 $
\nTherefore, y = $\\frac{2}{5}$ or $ \\frac{3}{4} $
\nTherefore, $x < y$
\nHence, option C is the correct answer.<\/p>\n
\n$ 8y^2-70y+153 = 0 $ can be written as $ (y-\\frac{9}{2})(y-\\frac{17}{4}) =0 $
\nTherefore, y = $\\frac{9}{2} $or $ \\frac{17}{4} $
\nTherefore, $x < y$
\nHence, option C is the correct answer.<\/p>\n
\n$ 3y^2-16y+5 = 0 $ can be written as $ (y-\\frac{1}{3})(y-5) =0 $
\nTherefore, y = $\\frac{1}{3}$ or $ 5 $
\nTherefore, No relation can be established
\nHence, option E is the correct answer.<\/p>\n
\n$ 10y^2-17y+3 = 0 $ can be written as $ (y-\\frac{1}{5})(y-\\frac{3}{2}) =0 $
\nTherefore, y = $\\frac{3}{2}$ or $ \\frac{1}{5} $
\nTherefore, no relation can be established
\nHence, option B is the correct answer.<\/p>\n
\n$2y^2-7y+6 = 0$ can be written as $(y-\\frac{3}{2})(y-2) = 0$
\ni.e. y = $\\frac{3}{2}$ or $2$
\nHence, $x>y$
\nHence, option A is the correct answer.<\/p>\n
\n$6y^2-11y+4 = 0$ can be written as $(y-\\frac{4}{3})(y-\\frac{1}{2}) = 0$
\ni.e. y = $\\frac{4}{3} or \\frac{1}{2}$
\nHence, no relation can be established
\nHence, option E is the correct answer.<\/p>\n
\n$12y^2-11y-5 = 0$ can be written as $(y-\\frac{5}{4})(y+\\frac{1}{3}) = 0$
\ni.e. y = $\\frac{5}{4} or -\\frac{1}{3}$
\nHence, no relation can be established.
\nHence, option E is the correct answer.<\/p>\n
\n$8y^2-14y+5 = 0$ can be written as $(y-\\frac{5}{4})(y-\\frac{1}{2}) = 0$
\ni.e. y = $\\frac{5}{4} or \\frac{1}{2}$
\nHence, $x > y$
\nHence, option A is the correct answer.<\/p>\n
\n$12y^2-19y+5 = 0$ can be written as $(y-\\frac{1}{3})(y-\\frac{5}{4}) = 0$
\ni.e. y = $\\frac{1}{3} or \\frac{5}{4}$
\nHence, No relation can be established
\nHence, option E is the correct answer.<\/p>\n
\ni.e. x = $\\frac{5}{4} or \\frac{1}{3}$
\n$20y^2-57y+40 = 0$ can be written as $(y-\\frac{5}{4})(y-\\frac{8}{5}) = 0$
\ni.e. y = $\\frac{5}{4} or \\frac{8}{5}$
\nHence, $x\\leq y$
\nHence, option D is the correct answer.<\/p>\n
\n$12y^2-19y+5 = 0$ can be written as $(y-\\frac{1}{3})(y-\\frac{5}{4}) = 0$
\ni.e. y = $\\frac{1}{3} or \\frac{5}{4}$
\nHence, no relation can be established.
\nHence, option E is the correct answer.<\/p>\n
\ni.e. x = $\\frac{1}{5} or 2$
\n$21y^2+4y-1 = 0$ can be written as $(y-\\frac{1}{7})(y+\\frac{1}{3}) = 0$
\ni.e. y = $\\frac{1}{7} or -\\frac{1}{3}$
\nHence, $x>y$
\nHence, option A is the correct answer.<\/p>\n