{"id":38846,"date":"2019-12-05T16:34:00","date_gmt":"2019-12-05T11:04:00","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=38846"},"modified":"2019-12-05T16:34:23","modified_gmt":"2019-12-05T11:04:23","slug":"mathematical-inequalities-questions-for-ibps-clerk-set-2-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/mathematical-inequalities-questions-for-ibps-clerk-set-2-pdf\/","title":{"rendered":"Mathematical Inequalities Questions for IBPS-Clerk Set-2 PDF"},"content":{"rendered":"

Mathematical Inequalities Questions for IBPS-Clerk Set-2 PDF<\/span><\/strong><\/span><\/h1>\n

Download important Mathematical Inequalities 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 Set-2 PDF<\/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

<p “=””>In these questions two equations numbered I and II are given. You have to solve both the equations and give answer<\/p>\n

a: if x < y
\nb: if x > y
\nc: if x = y
\nd: if x = y
\ne: if x= y or relationship between x and y cannot be established<\/p>\n

Question 1:\u00a0<\/b>I. $2x^{2} + 23x + 63 = 0$
\nII. $4y^{2} + 19y + 21 = 0$<\/p>\n

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

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

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

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

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

Question 2:\u00a0<\/b>I. $3x^{2}+ 29x + 56 = 0$
\nII. $2y^{2} + 15y + 25 = 0$<\/p>\n

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

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

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

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

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

Question 3:\u00a0<\/b>I. $3x^{2} + 23x+ 44 = 0$
\nII. $3y^{2} + 20y + 33 = 0$<\/p>\n

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

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

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

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

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

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Question 4:\u00a0<\/b>I. $4x^{2} – 29x + 45 = 0$
\nII. $3y^{2} – 19y + 28 = 0$<\/p>\n

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

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

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

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

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

Question 5:\u00a0<\/b>I. $2x^{2} – 13x + 21 = 0$
\nII. $5y^{2} – 22y + 21 =0$<\/p>\n

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

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

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

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

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

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

<\/del>In these questions, two equations numbered I and II are given. You have to solve both the equations and mark the appropriate answer.<\/p>\n

Give answer :<\/p>\n

(1) If x < y<\/p>\n

(2) If x > y<\/p>\n

(3) If x \u2264 y<\/p>\n

(4) If x \u2265 y<\/p>\n

(5) If relationship between x and y cannot be determined<\/p>\n

Question 6:\u00a0<\/b>I. $x^2 – 9x + 18 = 0$
\nII. $5y^2 – 22y + 24 = 0$<\/p>\n

a)\u00a0If x < y<\/p>\n

b)\u00a0If x > y<\/p>\n

c)\u00a0If x \u2264 y<\/p>\n

d)\u00a0If x \u2265 y<\/p>\n

e)\u00a0If relationship between x and y cannot be determined<\/p>\n

IBPS Clerk Previous Papers<\/a><\/p>\n

Question 7:\u00a0<\/b>I. $6x^2 + 11x + 5 = 0$
\nII. $2y^2 + 5y + 3 = 0$<\/p>\n

a)\u00a0If x < y<\/p>\n

b)\u00a0If x > y<\/p>\n

c)\u00a0If x \u2264 y<\/p>\n

d)\u00a0If x \u2265 y<\/p>\n

e)\u00a0If relationship between x and y cannot be determined<\/p>\n

Question 8:\u00a0<\/b>I. $x^2 + 10x + 24 = 0$
\nII. $y^{2}-\\sqrt{625}=0$<\/p>\n

a)\u00a0If x < y<\/p>\n

b)\u00a0If x > y<\/p>\n

c)\u00a0If x \u2264 y<\/p>\n

d)\u00a0If x \u2265 y<\/p>\n

e)\u00a0If relationship between x and y cannot be determined<\/p>\n

Question 9:\u00a0<\/b>I. $10x^2 + 11x + 1 =0$
\nII. $15y^2 + 8y + 1 = 0$<\/p>\n

a)\u00a0If x < y<\/p>\n

b)\u00a0If x > y<\/p>\n

c)\u00a0If x \u2264 y<\/p>\n

d)\u00a0If x \u2265 y<\/p>\n

e)\u00a0If relationship between x and y cannot be determined<\/p>\n

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Question 10:\u00a0<\/b>I. 15x^2 – 11x + 2 =0
\nII. 10y^2 – 9y + 2 =0<\/p>\n

a)\u00a0If x < y<\/p>\n

b)\u00a0If x > y<\/p>\n

c)\u00a0If x \u2264 y<\/p>\n

d)\u00a0If x \u2265 y<\/p>\n

e)\u00a0If relationship between x and y cannot be determined<\/p>\n

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

In these questions, two equations numbered I and II are given. You have to solve both the equations and select the appropriate option.<\/p>\n

Question 11:\u00a0<\/b>I. $x^{2}=144$
\nII. $y^{2}-24y+144=0$<\/p>\n

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

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

c)\u00a0relationship between x and y cannot be determined<\/p>\n

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

e)\u00a0x> y<\/p>\n

Question 12:\u00a0<\/b>I. $2x^{2}-9x+10=0$
\nII. $2y^{2}-13y+20=0$<\/p>\n

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

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

c)\u00a0relationship between x and y cannot be determined<\/p>\n

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

e)\u00a0x> y<\/p>\n

Question 13:\u00a0<\/b>I. $2x^{2}+15x+27=0$
\nII. $2y^{2}+7y+6=0$<\/p>\n

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

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

c)\u00a0relationship between x and y cannot be determined<\/p>\n

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

e)\u00a0x> y<\/p>\n

Question 14:\u00a0<\/b>I. $3x^{2}-13x+12=0$
\nII.$3y^{2}-13y+14=0$<\/p>\n

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

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

c)\u00a0relationship between x and y cannot be determined<\/p>\n

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

e)\u00a0x > y<\/p>\n

Question 15:\u00a0<\/b>I. $5x^{2}+8x+3=0$
\nII. $3y^{2}+7y+4=0$<\/p>\n

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

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

c)\u00a0relationship between x and y cannot be determined<\/p>\n

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

e)\u00a0x > y<\/p>\n

Daily Free Banking Online Tests<\/a><\/p>\n

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

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

$2x^{2} + 23x + 63 = 0$.
\n$x=\\frac{-23\\pm\\sqrt{23^{2}-4\\times63\\times2}}{2\\times2}$
\n$x=\\frac{-23\\pm5}{4}$.
\n$x=\\frac{-9}{2},-7$.
\n$4y^{2} + 19y + 21 =0$.
\n<\/span>$y=\\frac{-19\\pm\\sqrt{19^{2}-4\\times21\\times4}}{2\\times4}$.
\n$y=\\frac{-19\\pm5}{8}$.
\n$y=-3,\\frac{(-7)}{4}$.
\nClearly,
\nx < y
\nHence, Option A is correct.<\/p>\n

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

$3x^{2} + 29x + 56 = 0$.
\n$x=\\frac{-29\\pm\\sqrt{29^{2}-4\\times56\\times3}}{2\\times3}$
\n$x=\\frac{-29\\pm13}{6}$.
\n$x=\\frac{-8}{3},-7$.
\n$2y^{2} + 15y + 25 =0$.
\n<\/span>$y=\\frac{-15\\pm\\sqrt{15^{2}-4\\times25\\times2}}{2\\times2}$.
\n$y=\\frac{-15\\pm5}{4}$.
\n$y=-5,\\frac{(-5)}{2}$.
\nClearly,
\nx < y
\nHence, Option A is correct.<\/p>\n

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

$3x^{2} + 23x + 44 = 0$.
\n$x=\\frac{-23\\pm\\sqrt{23^{2}-4\\times44\\times3}}{2\\times3}$
\n$x=\\frac{-23\\pm1}{6}$.
\n$x=\\frac{-11}{3},-4$.
\n$3y^{2} + 20y + 33 =0$.
\n<\/span>$y=\\frac{-20\\pm\\sqrt{20^{2}-4\\times33\\times3}}{2\\times3}$.
\n$y=\\frac{-20\\pm2}{6}$.
\n$y=-3,\\frac{(-11)}{3}$.
\nClearly,
\nx <= y
\nHence, Option D is correct.<\/p>\n

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

$4x^{2} – 29x + 45 = 0$.
\n$x=\\frac{29\\pm\\sqrt{29^{2}-4\\times45\\times4}}{2\\times4}$
\n$x=\\frac{29\\pm3}{8}$.
\n$x=\\frac{13}{4},4$.
\n$3y^{2} – 19y + 28 =0$.
\n<\/span>$y=\\frac{19\\pm\\sqrt{19^{2}-4\\times28\\times3}}{2\\times3}$.
\n$y=\\frac{19\\pm5}{6}$.
\n$y=4,\\frac{7}{3}$.
\nClearly,
\nif x= y or relationship between x and y cannot be established
\nHence, Option E is correct.<\/p>\n

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

$2x^{2} – 13x + 21 = 0$.
\n$x=\\frac{13\\pm\\sqrt{13^{2}-4\\times21\\times2}}{2\\times2}$
\n$x=\\frac{13\\pm1}{4}$.
\n$x=\\frac{7}{2},3$.
\n$5y^{2} – 22y + 21 =0$.
\n<\/span><\/span>$y=\\frac{22\\pm\\sqrt{22^{2}-4\\times21\\times5}}{2\\times5}$
\n$y=\\frac{22\\pm8}{10}$.
\n$y=3,\\frac{7}{5}$.
\nClearly,
\nx => y
\nHence, Option C is correct.
\n<\/span><\/p>\n

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

$x^2-9x+18 = 0$
\n$(x-3)(x-6) = 0$
\n$x = 3, 6$<\/p>\n

$5y^2-22y+24 = 0$
\n$(5y-12)(y-2) = 0$
\n$y = 2, \\frac{12}{5}$<\/p>\n

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

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

$6x^2+11x+5 = 0$
\n$(6x+5)(x+1) = 0$
\n$x = -1, -\\frac{5}{6}$<\/p>\n

$2y^2+5y+3 = 0$
\n$(2y+3)(y+1) = 0$
\n$y = -\\frac{3}{2}, -1$<\/p>\n

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

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

$x^2+10x+24 = 0$
\n$(x+4)(x+6) = 0$
\n$x = -4, -6$<\/p>\n

$y^2 = \\sqrt{625}$
\n$y^2 = 25$
\n$y = 5, -5$<\/p>\n

relationship between x and y cannot be established<\/p>\n

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

$10x^2+11x+1 = 0$
\n$(10x+1)(x+1) = 0$
\n$x = -1, -\\frac{1}{10}$<\/p>\n

$15y^2+8y+1 = 0$
\n$(5y+1)(3y+1) = 0$
\n$y = -\\frac{1}{5}, -\\frac{1}{3}$<\/p>\n

relationship between x and y cannot be established<\/p>\n

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

I. 15x^2 – 11x + 2 =0
\nTo find the roots of this equation use the formula\u00a0 x1 = (-b +\u221a(b^2 – 4ac))\/2a ; substitute a = 15, b = -11, c = 2 we get x1 = 2\/5 = 0.4
\nSimilarly to find the another root x2 =\u00a0 (-b -\u221a(b^2 – 4ac))\/2a; we get x2 = 1\/3 =0.33
\n(x1, x2) = (0.4, 0.33)<\/p>\n

II. 10y^2 – 9y + 2 =0
\nSimilarly using the above formula we find the roots for 10y^2 – 9y + 2 =0 (a = 10, b = -9, c = 2)
\n(y1, y2) = (0.5, 0.4)<\/p>\n

Comparing the roots (x1, x2) = (0.4, 0.33) with y1 =0.5
\nclearly y1 is greater than both x1 and x2.<\/p>\n

Now compare the roots (x1, x2) with y2 = 0.4
\nHere we can observe x1 = y2 = 0.4<\/p>\n

but x2<y2 i.e.,0.33<0.4<\/p>\n

In all cases x<y except one case where x=y. So the answer should be x \u2264 y.
\nHence option ‘C’.<\/p>\n

Get 25 IBPS Clerk mocks for Rs. 149. Enroll here<\/a><\/p>\n

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

$x^2 = 144$
\n$x = -12, 12$<\/p>\n

$y^-24y+144 = 0$
\n${(y-12)}^2 = 0$
\n$y =12$<\/p>\n

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

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

$2x^2-9x+10 = 0$
\n$(2x-5)(x-2) = 0$
\n$x = 2, \\frac{5}{2}$<\/p>\n

$2y^2-13y+20 = 0$
\n$(2y-5)(y-4) = 0$
\n$y = \\frac{5}{2}, 4$<\/p>\n

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

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

$2x^2+15x+27 = 0$
\n$(2x+9)(x+3) = 0$
\n$x = -\\frac{9}{2}, -3$<\/p>\n

$2y^2+7y+6 = 0$
\n$(2y+3)(y+2) = 0$
\n$y = -\\frac{3}{2}, -2$<\/p>\n

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

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

$3x^2-13x+12 = 0$
\n$(3x-4)(x-3) = 0$
\n$x = \\frac{4}{3}, 3$<\/p>\n

$3y^2-13y+14 = 0$
\n$(3y-7)(y-2) = 0$
\n$y = 2, \\frac{7}{3}$<\/p>\n

relationship between x and y cannot be established<\/p>\n

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

$5x^2+8x+3 = 0$
\n$(5x+3)(x+1) = 0$
\n$x = -1, -\\frac{3}{5}$<\/p>\n

$3y^2+7y+4 = 0$
\n$(3y+4)(y+1) = 0$
\n$y = -1, -\\frac{4}{3}$<\/p>\n

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

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We hope this Mathematical Inequalities for IBPS Clerk preparation will be helpful to you.<\/p>\n","protected":false},"excerpt":{"rendered":"

Mathematical Inequalities Questions for IBPS-Clerk Set-2 PDF Download important Mathematical Inequalities PDF based on previously asked questions in IBPS Clerk and other Banking Exams. Practice Mathematical Inequalities for IBPS Clerk Exam. Take Free IBPS Clerk Mock Test Download IBPS Clerk Previous papers PDF Go to Free Banking Study Material (15,000 Solved Questions) Instructions <p “=””>In 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