Mathematical Inequalities Questions For IBPS Clerk

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mathematical inequalities questions for ibps clerk
mathematical inequalities questions for ibps clerk

Mathematical Inequalities Questions For IBPS Clerk

Download important Mathematical Iequalities PDF based on previously asked questions in IBPS Clerk and other Banking Exams. Practice Mathematical Inequalities for IBPS Clerk Exam.

Download Mathematical Inequalities Questions For IBPS Clerk

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Instructions

In the following questions two equations numbered I and II are given. You have to solve both the equations and
Give answer a: if x > y
Give answer b: if x ≥ y
Give answer c: if x < y
Give answer d: if x ≤ y
Give answer e: if x = y or the relationship cannot be established.

Question 1: I.  $x^{2}-3x-88=0$
II. $y^{2}+8y-48=0$

a) if x > y

b) if x ≥ y

c) if x < y

d) if x ≤ y

e) if x = y or the relationship cannot be established.

Question 2: I.  $5x^{2}+29x+20=0$
II. $25y^{2}+25y+6=0$

a) if x > y

b) if x ≥ y

c) if x < y

d) if x ≤ y

e) if x = y or the relationship cannot be established.

Question 3: I.  $2x^{2}-11x+12=0$
II. $2y^{2}-19y+44=0$

a) if x > y

b) if x ≥ y

c) if x < y

d) if x ≤ y

e) if x = y or the relationship cannot be established.

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Question 4: I.  $3x^{2}+10x+8=0$
II. $3y^{2}+7y+4=0$

a) if x > y

b) if x ≥ y

c) if x < y

d) if x ≤ y

e) if x = y or the relationship cannot be established.

Question 5: I.  $2x^{2}+21x+10=0$
II. $3y^{2}+13y+14=0$

a) if x > y

b) if x ≥ y

c) if x < y

d) if x ≤ y

e) if x = y or the relationship cannot be established.

Instructions

In each of these questions two equations numbered I and II are given. You have to solve both the equations and –
Give answer a: if x < y
Give answer b: if x ≤ y
Give answer c: if x > y
Give answer d: if x ≥ y
Give answer e: if x = y or the relationship cannot be established.

Question 6: I.  $x^{2}+13x+42=0$
II. $y^{2} +19y+90=0$

a) if x < y

b) if x ≤ y

c) if x > y

d) if x ≥ y

e) if x = y or the relationship cannot be established.

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Question 7: I.   $x^{2}-15x+56=0$
II. $y^{2} -23y+132=0$

a) if x < y

b) if x ≤ y

c) if x > y

d) if x ≥ y

e) if x = y or the relationship cannot be established.

Question 8: I. $x^{2}+7x+12=0$
II. $y^{2} +6y+8=0$

a) if x < y

b) if x ≤ y

c) if x > y

d) if x ≥ y

e) if x = y or the relationship cannot be established.

Question 9: I. $x^{2}-22x+120=0$
II. $y^{2} -26y+168=0$

a) if x < y

b) if x ≤ y

c) if x > y

d) if x ≥ y

e) if x = y or the relationship cannot be established.

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Question 10: I.$x^{2}+12x+32=0$
II. $y^{2} +17y+72=0$

a) if x < y

b) if x ≤ y

c) if x > y

d) if x ≥ y

e) if x = y or the relationship cannot be established.

Instructions

In each of the following questions, two equations I and II have been given.
Solve these questions and answer

(1)if x < y
(2) if x ≤ y

(3) if x = y or the relation cannot be established

(4) if ≥ y

(5) if x > y

Question 11: I. $30 x^{2} + 11x + 1 = 0$
II. $42 y^{2} + 13y + 1 = 0$

a) if x < y

b) if x ≤ y

c) if x = y or the relation cannot be established

d)  if ≥ y

e) if x > y

Question 12: I. $x^{2}- x – \sqrt{2}x + \sqrt{2}=0$
II.$y^{2}-3y+2=0$

a) if x < y

b)  if x ≤ y

c) if x = y or the relation cannot be established

d) if ≥ y

e) if x > y

Question 13: I.$x^{2}-2x-\sqrt{5}x+2\sqrt{5}=0$
II.$y^{2}-\sqrt{3}y-\sqrt{2}y+\sqrt{6}=0$

a) if x < y

b)  if x ≤ y

c) if x = y or the relation cannot be established

d) if ≥ y

e) if x > y

Question 14: I.$ x^{2}+12x+36=0$
II.$y^{2}=16$

a) if x < y

b) if x ≤ y

c) if x = y or the relation cannot be established

d)  if ≥ y

e) if x > y

Question 15: I.$9x^{2}+3x-2=0$
II.$8y^{2}+6y+1=0$

a) if x < y

b)  if x ≤ y

c) if x = y or the relation cannot be established

d) if ≥ y

e) if x > y

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Answers & Solutions:

1) Answer (E)

I.$x^{2} – 3x – 88 = 0$

=> $x^2 + 8x – 11x – 88 = 0$

=> $x (x + 8) – 11 (x + 8) = 0$

=> $(x + 8) (x – 11) = 0$

=> $x = -8 , 11$

II.$y^{2} + 8y – 48 = 0$

=> $y^2 + 12y – 4y – 48 = 0$

=> $y (y + 12) – 4 (y + 12) = 0$

=> $(y + 12) (y – 4) = 0$

=> $y = -12 , 4$

$\therefore$ No relation can be established.

2) Answer (C)

I.$5x^{2} + 29x + 20 = 0$

=> $5x^2 + 25x + 4x + 20 = 0$

=> $5x (x + 5) + 4 (x + 5) = 0$

=> $(x + 5) (5x + 4) = 0$

=> $x = -5 , \frac{-4}{5}$

II.$25y^{2} + 25y + 6 = 0$

=> $25y^2 + 10y + 15y + 6 = 0$

=> $5y (5y + 2) + 3 (5y + 2) = 0$

=> $(5y + 3) (5y + 2) = 0$

=> $y = \frac{-3}{5} , \frac{-2}{5}$

Therefore $x < y$

3) Answer (D)

I.$2x^{2} – 11x + 12 = 0$

=> $2x^2 – 8x – 3x + 12 = 0$

=> $2x (x – 4) – 3 (x – 4) = 0$

=> $(x – 4) (2x – 3) = 0$

=> $x = 4 , \frac{3}{2}$

II.$2y^{2} – 19y + 44 = 0$

=> $2y^2 – 8y – 11y + 44 = 0$

=> $2y (y – 4) – 11 (y – 4) = 0$

=> $(y – 4) (2y – 11) = 0$

=> $y = 4 , \frac{11}{2}$

$\therefore x \leq y$

4) Answer (D)

I.$3x^{2} + 10x + 8 = 0$

=> $3x^2 + 6x + 4x + 8 = 0$

=> $3x (x + 2) + 4 (x + 2) = 0$

=> $(x + 2) (3x + 4) = 0$

=> $x = -2 , \frac{-4}{3}$

II.$3y^{2} + 7y + 4 = 0$

=> $3y^2 + 3y + 4y + 4 = 0$

=> $3y (y + 1) + 4 (y + 1) = 0$

=> $(y + 1) (3y + 4) = 0$

=> $y = -1 , \frac{-4}{3}$

$\therefore x \leq y$

5) Answer (E)

I.$2x^{2} + 21x + 10 = 0$

=> $2x^2 + x + 20x + 10 = 0$

=> $x (2x + 1) + 10 (2x + 1) = 0$

=> $(x + 10) (2x + 1) = 0$

=> $x = -10 , \frac{-1}{2}$

II.$3y^{2} + 13y + 14 = 0$

=> $3y^2 + 6y + 7y + 14 = 0$

=> $3y (y + 2) + 7 (y + 2) = 0$

=> $(y + 2) (3y + 7) = 0$

=> $y = -2 , \frac{-7}{3}$

$\therefore$ No relation can be established.

6) Answer (C)

I.$x^{2} + 13x + 42 = 0$

=> $x^2 + 7x + 6x + 42 = 0$

=> $x (x + 7) + 6 (x + 7) = 0$

=> $(x + 7) (x + 6) = 0$

=> $x = -7 , -6$

II.$y^{2} + 19y + 90 = 0$

=> $y^2 + 9y + 10y + 90 = 0$

=> $y (y + 9) + 10 (y + 9) = 0$

=> $(y + 9) (y + 10) = 0$

=> $y = -9 , -10$

$\therefore x > y$

7) Answer (A)

I.$x^{2} – 15x + 56 = 0$

=> $x^2 – 8x – 7x + 56 = 0$

=> $x (x – 8) – 7 (x – 8) = 0$

=> $(x – 8) (x – 7) = 0$

=> $x = 8 , 7$

II.$y^{2} – 23y + 132 = 0$

=> $y^2 – 11y – 12y + 132 = 0$

=> $y (y – 11) – 12 (y – 11) = 0$

=> $(y – 11) (y – 12) = 0$

=> $y = 11 , 12$

$\therefore x < y$

8) Answer (E)

I.$x^{2} + 7x + 12 = 0$

=> $x^2 + 3x + 4x + 12 = 0$

=> $x (x + 3) + 4 (x + 3) = 0$

=> $(x + 3) (x + 4) = 0$

=> $x = -3 , -4$

II.$y^{2} + 6y + 8 = 0$

=> $y^2 + 4y + 2y + 8 = 0$

=> $y (y + 4) + 2 (y + 4) = 0$

=> $(y + 4) (y + 2) = 0$

=> $y = -4 , -2$

Because $-2 > -4$ and $-3 > -4$

Therefore, no relation can be established.

9) Answer (B)

I.$x^{2} – 22x + 120 = 0$

=> $x^2 – 10x – 12x + 120 = 0$

=> $x (x – 10) – 12 (x – 10) = 0$

=> $(x – 10) (x – 12) = 0$

=> $x = 10 , 12$

II.$y^{2} – 26y + 168 = 0$

=> $y^2 – 12y – 14y + 168 = 0$

=> $y (y – 12) – 14 (y – 12) = 0$

=> $(y – 12) (y – 14) = 0$

=> $y = 12 , 14$

$\therefore x \leq y$

10) Answer (D)

I.$x^{2} + 12x + 32 = 0$

=> $x^2 + 8x + 4x + 32 = 0$

=> $x (x + 8) + 4 (x + 8) = 0$

=> $(x + 8) (x + 4) = 0$

=> $x = -8 , -4$

II.$y^{2} + 17y + 72 = 0$

=> $y^2 + 9y + 8y + 72 = 0$

=> $y (y + 9) + 8 (y + 9) = 0$

=> $(y + 9) (y + 8) = 0$

=> $y = -9 , -8$

$\therefore x \geq y$

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11) Answer (B)

Statement I : $30 x^{2} + 11x + 1 = 0$

=> $30x^2 + 6x + 5x + 1 = 0$

=> $6x (5x + 1) + 1 (5x + 1) = 0$

=> $(6x + 1) (5x + 1) = 0$

=> $x = \frac{-1}{6} , \frac{-1}{5}$

Statement II : $42 y^{2} + 13y + 1 = 0$

=> $42y^2 + 7y + 6y + 1 = 0$

=> $7y (6y + 1) + 1 (6y + 1) = 0$

=> $(7y + 1) (6y + 1) = 0$

=> $y = \frac{-1}{7} , \frac{-1}{6}$

$\therefore$ $x \leq y$

12) Answer (C)

I. $x^{2}- x – \sqrt{2}x + \sqrt{2}=0$

=> $x (x – 1) – \sqrt{2} (x – 1) = 0$

=> $(x – \sqrt{2}) (x – 1) = 0$

=> $x = \sqrt{2} , 1$

II. $y^{2}-3y+2=0$

=> $y^2 – 2y – y + 2 = 0$

=> $y (y – 2) – 1 (y – 2) = 0$

=> $(y – 2) (y – 1) = 0$

=> $y = 1 , 2$

$\therefore$ No relation established.

13) Answer (E)

Statement I : $x^{2}-2x-\sqrt{5}x+2\sqrt{5}=0$

=> $x (x – 2) – \sqrt{5} (x – 2) = 0$

=> $(x – \sqrt{5}) (x – 2) = 0$

=> $x = \sqrt{5} , 2$

Statement II : $y^{2}-\sqrt{3}y-\sqrt{2}y+\sqrt{6}=0$

=> $y (y – \sqrt{3}) – \sqrt{2} (y – \sqrt{3}) = 0$

=> $(y – \sqrt{2}) (y – \sqrt{3}) = 0$

=> $y = \sqrt{2} , \sqrt{3}$

$\therefore$ $x > y$

14) Answer (A)

Statement 1 : $x^2 + 12x + 36 = 0$

=> $x^2 + 2.x.6 + 6^2 = 0$

=> $(x + 6)^2 = 0$

=> $x = -6$

Statement II : $y^2 = 16$

=> $(y)^2 = (\pm 4)^2$

=> $y = \pm 4$

$\therefore$ $x < y$

15) Answer (C)

Statement I : $9x^{2}+3x-2=0$

=> $9x^2 + 6x – 3x – 2 = 0$

=> $3x (3x + 2) – 1 (3x + 2) = 0$

=> $(3x – 1) (3x + 2) = 0$

=> $x = \frac{1}{3} , \frac{-2}{3}$

Statement II : $8y^{2}+6y+1=0$

=> $8y^2 + 4y + 2y + 1 = 0$

=> $4y (2y + 1) + 1 (2y + 1) = 0$

=> $(4y + 1) (2y + 1) = 0$

=> $y = \frac{-1}{4} , \frac{-1}{2}$

$\therefore$ No relation can be established.

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