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# 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.

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Instructions

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

a: if x < y
b: if x > y
c: if x = y
d: if x = y
e: if x= y or relationship between x and y cannot be established

Question 1: I. $2x^{2} + 23x + 63 = 0$
II. $4y^{2} + 19y + 21 = 0$

a) if x < y

b) if x > y

c) if x => y

d) if x <= y

e) if x= y or relationship between x and y cannot be established

Question 2: I. $3x^{2}+ 29x + 56 = 0$
II. $2y^{2} + 15y + 25 = 0$

a) if x < y

b) if x > y

c) if x => y

d) if x <= y

e) if x= y or relationship between x and y cannot be established

Question 3: I. $3x^{2} + 23x+ 44 = 0$
II. $3y^{2} + 20y + 33 = 0$

a) if x < y

b) if x > y

c) if x => y

d) if x <= y

e) if x= y or relationship between x and y cannot be established

Question 4: I. $4x^{2} – 29x + 45 = 0$
II. $3y^{2} – 19y + 28 = 0$

a) if x < y

b) if x > y

c) if x => y

d) if x <= y

e) if x= y or relationship between x and y cannot be established

Question 5: I. $2x^{2} – 13x + 21 = 0$
II. $5y^{2} – 22y + 21 =0$

a) if x < y

b) if x > y

c) if x => y

d) if x <= y

e) if x= y or relationship between x and y cannot be established

Instructions

In these questions, two equations numbered I and II are given. You have to solve both the equations and mark the appropriate answer.

(1) If x < y

(2) If x > y

(3) If x ≤ y

(4) If x ≥ y

(5) If relationship between x and y cannot be determined

Question 6: I. $x^2 – 9x + 18 = 0$
II. $5y^2 – 22y + 24 = 0$

a) If x < y

b) If x > y

c) If x ≤ y

d) If x ≥ y

e) If relationship between x and y cannot be determined

Question 7: I. $6x^2 + 11x + 5 = 0$
II. $2y^2 + 5y + 3 = 0$

a) If x < y

b) If x > y

c) If x ≤ y

d) If x ≥ y

e) If relationship between x and y cannot be determined

Question 8: I. $x^2 + 10x + 24 = 0$
II. $y^{2}-\sqrt{625}=0$

a) If x < y

b) If x > y

c) If x ≤ y

d) If x ≥ y

e) If relationship between x and y cannot be determined

Question 9: I. $10x^2 + 11x + 1 =0$
II. $15y^2 + 8y + 1 = 0$

a) If x < y

b) If x > y

c) If x ≤ y

d) If x ≥ y

e) If relationship between x and y cannot be determined

Question 10: I. 15x^2 – 11x + 2 =0
II. 10y^2 – 9y + 2 =0

a) If x < y

b) If x > y

c) If x ≤ y

d) If x ≥ y

e) If relationship between x and y cannot be determined

Instructions

In these questions, two equations numbered I and II are given. You have to solve both the equations and select the appropriate option.

Question 11: I. $x^{2}=144$
II. $y^{2}-24y+144=0$

a) x ≤ y

b) x ≥ y

c) relationship between x and y cannot be determined

d) x< y

e) x> y

Question 12: I. $2x^{2}-9x+10=0$
II. $2y^{2}-13y+20=0$

a) x ≤ y

b) x ≥ y

c) relationship between x and y cannot be determined

d) x< y

e) x> y

Question 13: I. $2x^{2}+15x+27=0$
II. $2y^{2}+7y+6=0$

a) x ≤ y

b) x ≥ y

c) relationship between x and y cannot be determined

d) x < y

e) x> y

Question 14: I. $3x^{2}-13x+12=0$
II.$3y^{2}-13y+14=0$

a) x≤ y

b) x ≥ y

c) relationship between x and y cannot be determined

d) x < y

e) x > y

Question 15: I. $5x^{2}+8x+3=0$
II. $3y^{2}+7y+4=0$

a) x ≤ y

b) x ≥ y

c) relationship between x and y cannot be determined

d) x < y

e) x > y

$2x^{2} + 23x + 63 = 0$.
$x=\frac{-23\pm\sqrt{23^{2}-4\times63\times2}}{2\times2}$
$x=\frac{-23\pm5}{4}$.
$x=\frac{-9}{2},-7$.
$4y^{2} + 19y + 21 =0$.
$y=\frac{-19\pm\sqrt{19^{2}-4\times21\times4}}{2\times4}$.
$y=\frac{-19\pm5}{8}$.
$y=-3,\frac{(-7)}{4}$.
Clearly,
x < y
Hence, Option A is correct.

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

$3x^{2} + 23x + 44 = 0$.
$x=\frac{-23\pm\sqrt{23^{2}-4\times44\times3}}{2\times3}$
$x=\frac{-23\pm1}{6}$.
$x=\frac{-11}{3},-4$.
$3y^{2} + 20y + 33 =0$.
$y=\frac{-20\pm\sqrt{20^{2}-4\times33\times3}}{2\times3}$.
$y=\frac{-20\pm2}{6}$.
$y=-3,\frac{(-11)}{3}$.
Clearly,
x <= y
Hence, Option D is correct.

$4x^{2} – 29x + 45 = 0$.
$x=\frac{29\pm\sqrt{29^{2}-4\times45\times4}}{2\times4}$
$x=\frac{29\pm3}{8}$.
$x=\frac{13}{4},4$.
$3y^{2} – 19y + 28 =0$.
$y=\frac{19\pm\sqrt{19^{2}-4\times28\times3}}{2\times3}$.
$y=\frac{19\pm5}{6}$.
$y=4,\frac{7}{3}$.
Clearly,
if x= y or relationship between x and y cannot be established
Hence, Option E is correct.

$2x^{2} – 13x + 21 = 0$.
$x=\frac{13\pm\sqrt{13^{2}-4\times21\times2}}{2\times2}$
$x=\frac{13\pm1}{4}$.
$x=\frac{7}{2},3$.
$5y^{2} – 22y + 21 =0$.
$y=\frac{22\pm\sqrt{22^{2}-4\times21\times5}}{2\times5}$
$y=\frac{22\pm8}{10}$.
$y=3,\frac{7}{5}$.
Clearly,
x => y
Hence, Option C is correct.

$x^2-9x+18 = 0$
$(x-3)(x-6) = 0$
$x = 3, 6$

$5y^2-22y+24 = 0$
$(5y-12)(y-2) = 0$
$y = 2, \frac{12}{5}$

$x > y$

$6x^2+11x+5 = 0$
$(6x+5)(x+1) = 0$
$x = -1, -\frac{5}{6}$

$2y^2+5y+3 = 0$
$(2y+3)(y+1) = 0$
$y = -\frac{3}{2}, -1$

$x\geq y$

$x^2+10x+24 = 0$
$(x+4)(x+6) = 0$
$x = -4, -6$

$y^2 = \sqrt{625}$
$y^2 = 25$
$y = 5, -5$

relationship between x and y cannot be established

$10x^2+11x+1 = 0$
$(10x+1)(x+1) = 0$
$x = -1, -\frac{1}{10}$

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

relationship between x and y cannot be established

I. 15x^2 – 11x + 2 =0
To find the roots of this equation use the formula  x1 = (-b +√(b^2 – 4ac))/2a ; substitute a = 15, b = -11, c = 2 we get x1 = 2/5 = 0.4
Similarly to find the another root x2 =  (-b -√(b^2 – 4ac))/2a; we get x2 = 1/3 =0.33
(x1, x2) = (0.4, 0.33)

II. 10y^2 – 9y + 2 =0
Similarly using the above formula we find the roots for 10y^2 – 9y + 2 =0 (a = 10, b = -9, c = 2)
(y1, y2) = (0.5, 0.4)

Comparing the roots (x1, x2) = (0.4, 0.33) with y1 =0.5
clearly y1 is greater than both x1 and x2.

Now compare the roots (x1, x2) with y2 = 0.4
Here we can observe x1 = y2 = 0.4

but x2<y2 i.e.,0.33<0.4

In all cases x<y except one case where x=y. So the answer should be x ≤ y.
Hence option ‘C’.

$x^2 = 144$
$x = -12, 12$

$y^-24y+144 = 0$
${(y-12)}^2 = 0$
$y =12$

$x\leq y$

$2x^2-9x+10 = 0$
$(2x-5)(x-2) = 0$
$x = 2, \frac{5}{2}$

$2y^2-13y+20 = 0$
$(2y-5)(y-4) = 0$
$y = \frac{5}{2}, 4$

$x\leq y$

$2x^2+15x+27 = 0$
$(2x+9)(x+3) = 0$
$x = -\frac{9}{2}, -3$

$2y^2+7y+6 = 0$
$(2y+3)(y+2) = 0$
$y = -\frac{3}{2}, -2$

$x < y$

$3x^2-13x+12 = 0$
$(3x-4)(x-3) = 0$
$x = \frac{4}{3}, 3$

$3y^2-13y+14 = 0$
$(3y-7)(y-2) = 0$
$y = 2, \frac{7}{3}$

relationship between x and y cannot be established

$5x^2+8x+3 = 0$
$(5x+3)(x+1) = 0$
$x = -1, -\frac{3}{5}$
$3y^2+7y+4 = 0$
$(3y+4)(y+1) = 0$
$y = -1, -\frac{4}{3}$
$x\geq y$