Mathematical Inequalities Questions for IBPS-Clerk Set-2 PDF

0
5437
mathematical inequalities questions for ibps set-2 clerk pdf
mathematical inequalities questions for ibps set-2 clerk pdf

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.

Download Mathematical Inequalities Questions Set-2 PDF

Get 25 IBPS Clerk mocks for Rs. 149. Enroll here

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 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

IBPS Clerk Online Mock Test

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.

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: 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

IBPS Clerk Previous Papers

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

IBPS Clerk Important Questions PDF

Free Banking Study Material (15,000 Solved Questions)

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

Daily Free Banking Online Tests

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’.

Get 25 IBPS Clerk mocks for Rs. 149. Enroll here

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)

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

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

$x\geq y$

Highly Rated Free Preparation App for Banking Exams

4 Free IBPS Clerk Mock Tests

We hope this Mathematical Inequalities for IBPS Clerk preparation will be helpful to you.

LEAVE A REPLY

Please enter your comment!
Please enter your name here