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:Â I. $2x^{2} + 23x + 63 = 0$
II. $4y^{2} + 19y + 21 = 0$
a)Â if x < y
b)Â if x > y
c)Â if x => y
d)Â if x <= y
e)Â if x= y or relationship between x and y cannot be established
Question 2:Â I. $3x^{2}+ 29x + 56 = 0$
II. $2y^{2} + 15y + 25 = 0$
a)Â if x < y
b)Â if x > y
c)Â if x => y
d)Â if x <= y
e)Â if x= y or relationship between x and y cannot be established
Question 3:Â I. $3x^{2} + 23x+ 44 = 0$
II. $3y^{2} + 20y + 33 = 0$
a)Â if x < y
b)Â if x > y
c)Â if x => y
d)Â if x <= y
e)Â if x= y or relationship between x and y cannot be established
Question 4:Â I. $4x^{2} – 29x + 45 = 0$
II. $3y^{2} – 19y + 28 = 0$
a)Â if x < y
b)Â if x > y
c)Â if x => y
d)Â if x <= y
e)Â if x= y or relationship between x and y cannot be established
Question 5:Â I. $2x^{2} – 13x + 21 = 0$
II. $5y^{2} – 22y + 21 =0$
a)Â if x < y
b)Â if x > y
c)Â if x => y
d)Â if x <= y
e)Â 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.
Give 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:Â I. $x^2 – 9x + 18 = 0$
II. $5y^2 – 22y + 24 = 0$
a)Â If x < y
b)Â If x > y
c) If x ≤ y
d) If x ≥ y
e)Â If relationship between x and y cannot be determined
Question 7:Â I. $6x^2 + 11x + 5 = 0$
II. $2y^2 + 5y + 3 = 0$
a)Â If x < y
b)Â If x > y
c) If x ≤ y
d) If x ≥ y
e)Â If relationship between x and y cannot be determined
Question 8:Â I. $x^2 + 10x + 24 = 0$
II. $y^{2}-\sqrt{625}=0$
a)Â If x < y
b)Â If x > y
c) If x ≤ y
d) If x ≥ y
e)Â If relationship between x and y cannot be determined
Question 9:Â I. $10x^2 + 11x + 1 =0$
II. $15y^2 + 8y + 1 = 0$
a)Â If x < y
b)Â If x > y
c) If x ≤ y
d) If x ≥ y
e)Â If relationship between x and y cannot be determined
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Question 10:Â I. 15x^2 – 11x + 2 =0
II. 10y^2 – 9y + 2 =0
a)Â If x < y
b)Â If x > y
c) If x ≤ y
d) If x ≥ y
e)Â 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:Â I. $x^{2}=144$
II. $y^{2}-24y+144=0$
a) x ≤ y
b) x ≥ y
c)Â relationship between x and y cannot be determined
d)Â x< y
e)Â x> y
Question 12:Â I. $2x^{2}-9x+10=0$
II. $2y^{2}-13y+20=0$
a) x ≤ y
b) x ≥ y
c)Â relationship between x and y cannot be determined
d)Â x< y
e)Â x> y
Question 13:Â I. $2x^{2}+15x+27=0$
II. $2y^{2}+7y+6=0$
a) x ≤ y
b) x ≥ y
c)Â relationship between x and y cannot be determined
d)Â x < y
e)Â x> y
Question 14:Â I. $3x^{2}-13x+12=0$
II.$3y^{2}-13y+14=0$
a) x≤ y
b) x ≥ y
c)Â relationship between x and y cannot be determined
d)Â x < y
e)Â x > y
Question 15:Â I. $5x^{2}+8x+3=0$
II. $3y^{2}+7y+4=0$
a) x ≤ y
b) x ≥ y
c)Â relationship between x and y cannot be determined
d)Â x < y
e)Â x > y
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Answers & Solutions:
1) Answer (A)
$2x^{2} + 23x + 63 = 0$.
$x=\frac{-23\pm\sqrt{23^{2}-4\times63\times2}}{2\times2}$
$x=\frac{-23\pm5}{4}$.
$x=\frac{-9}{2},-7$.
$4y^{2} + 19y + 21 =0$.
$y=\frac{-19\pm\sqrt{19^{2}-4\times21\times4}}{2\times4}$.
$y=\frac{-19\pm5}{8}$.
$y=-3,\frac{(-7)}{4}$.
Clearly,
x < y
Hence, Option A is correct.
2) Answer (A)
$3x^{2} + 29x + 56 = 0$.
$x=\frac{-29\pm\sqrt{29^{2}-4\times56\times3}}{2\times3}$
$x=\frac{-29\pm13}{6}$.
$x=\frac{-8}{3},-7$.
$2y^{2} + 15y + 25 =0$.
$y=\frac{-15\pm\sqrt{15^{2}-4\times25\times2}}{2\times2}$.
$y=\frac{-15\pm5}{4}$.
$y=-5,\frac{(-5)}{2}$.
Clearly,
x < y
Hence, Option A is correct.
3) Answer (D)
$3x^{2} + 23x + 44 = 0$.
$x=\frac{-23\pm\sqrt{23^{2}-4\times44\times3}}{2\times3}$
$x=\frac{-23\pm1}{6}$.
$x=\frac{-11}{3},-4$.
$3y^{2} + 20y + 33 =0$.
$y=\frac{-20\pm\sqrt{20^{2}-4\times33\times3}}{2\times3}$.
$y=\frac{-20\pm2}{6}$.
$y=-3,\frac{(-11)}{3}$.
Clearly,
x <= y
Hence, Option D is correct.
4) Answer (E)
$4x^{2} – 29x + 45 = 0$.
$x=\frac{29\pm\sqrt{29^{2}-4\times45\times4}}{2\times4}$
$x=\frac{29\pm3}{8}$.
$x=\frac{13}{4},4$.
$3y^{2} – 19y + 28 =0$.
$y=\frac{19\pm\sqrt{19^{2}-4\times28\times3}}{2\times3}$.
$y=\frac{19\pm5}{6}$.
$y=4,\frac{7}{3}$.
Clearly,
if x= y or relationship between x and y cannot be established
Hence, Option E is correct.
5) Answer (C)
$2x^{2} – 13x + 21 = 0$.
$x=\frac{13\pm\sqrt{13^{2}-4\times21\times2}}{2\times2}$
$x=\frac{13\pm1}{4}$.
$x=\frac{7}{2},3$.
$5y^{2} – 22y + 21 =0$.
$y=\frac{22\pm\sqrt{22^{2}-4\times21\times5}}{2\times5}$
$y=\frac{22\pm8}{10}$.
$y=3,\frac{7}{5}$.
Clearly,
x => y
Hence, Option C is correct.
6) Answer (B)
$x^2-9x+18 = 0$
$(x-3)(x-6) = 0$
$x = 3, 6$
$5y^2-22y+24 = 0$
$(5y-12)(y-2) = 0$
$y = 2, \frac{12}{5}$
$x > y$
7) Answer (D)
$6x^2+11x+5 = 0$
$(6x+5)(x+1) = 0$
$x = -1, -\frac{5}{6}$
$2y^2+5y+3 = 0$
$(2y+3)(y+1) = 0$
$y = -\frac{3}{2}, -1$
$x\geq y$
8) Answer (E)
$x^2+10x+24 = 0$
$(x+4)(x+6) = 0$
$x = -4, -6$
$y^2 = \sqrt{625}$
$y^2 = 25$
$y = 5, -5$
relationship between x and y cannot be established
9) Answer (E)
$10x^2+11x+1 = 0$
$(10x+1)(x+1) = 0$
$x = -1, -\frac{1}{10}$
$15y^2+8y+1 = 0$
$(5y+1)(3y+1) = 0$
$y = -\frac{1}{5}, -\frac{1}{3}$
relationship between x and y cannot be established
10) Answer (C)
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’.
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11) Answer (A)
$x^2 = 144$
$x = -12, 12$
$y^-24y+144 = 0$
${(y-12)}^2 = 0$
$y =12$
$x\leq y$
12) Answer (A)
$2x^2-9x+10 = 0$
$(2x-5)(x-2) = 0$
$x = 2, \frac{5}{2}$
$2y^2-13y+20 = 0$
$(2y-5)(y-4) = 0$
$y = \frac{5}{2}, 4$
$x\leq y$
13) Answer (D)
$2x^2+15x+27 = 0$
$(2x+9)(x+3) = 0$
$x = -\frac{9}{2}, -3$
$2y^2+7y+6 = 0$
$(2y+3)(y+2) = 0$
$y = -\frac{3}{2}, -2$
$x < y$
14) Answer (C)
$3x^2-13x+12 = 0$
$(3x-4)(x-3) = 0$
$x = \frac{4}{3}, 3$
$3y^2-13y+14 = 0$
$(3y-7)(y-2) = 0$
$y = 2, \frac{7}{3}$
relationship between x and y cannot be established
15) Answer (B)
$5x^2+8x+3 = 0$
$(5x+3)(x+1) = 0$
$x = -1, -\frac{3}{5}$
$3y^2+7y+4 = 0$
$(3y+4)(y+1) = 0$
$y = -1, -\frac{4}{3}$
$x\geq y$
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