Mathematical Inequalities Questions for SBI Clerk PDF
Download SBI Clerk Mathematical Inequalities Questions & Answers PDF for SBI Clerk Prelims and Mains exam. Very Important SBI Clerk Mathematical Inequalities Questions with solutions.
Download Mathematical Inequalities Questions for SBI Clerk PDF
105 SBI Clerk mocks for Rs. 199. Enroll here
Instructions
For the two given equations I and II—-
Question 1: I. $6p^{2}+5p+1=0$
II. $20q^{2}+9q=-1$
a) Give answer (A) if p is greater than q.
b) Give answer (B) if p is smaller than q.
c) Give answer (C) if p is equal to q.
d) Give answer (D) if p is either equal to or greater than q.
e) Give answer (E) if p is either equal to or smaller than q.
Question 2: I. $3p^{2}+2p-1=0$ II. $2q^{2}+7q+6=0$
a) Give answer (A) if p is greater than q.
b) Give answer (B) if p is smaller than q.
c) Give answer (C) if p is equal to q.
d) Give answer (D) if p is either equal to or greater than q.
e) Give answer (E) if p is either equal to or smaller than q.
Question 3: I. $3p^2+15p=-18$ II. $q^2+7q+12=0$
a) Give answer (A) if p is greater than q.
b) Give answer (B) if p is smaller than q.
c) Give answer (C) if p is equal to q.
d) Give answer (D) if p is either equal to or greater than q.
e) Give answer (E) if p is either equal to or smaller than q.
Question 4: I. $p=\frac{\sqrt{4}}{\sqrt{9}}$ II. $9q^{2}-12q+4=0$
a) Give answer (A) if p is greater than q.
b) Give answer (B) if p is smaller than q.
c) Give answer (C) if p is equal to q.
d) Give answer (D) if p is either equal to or greater than q.
e) Give answer (E) if p is either equal to or smaller than q.
Question 5: I. $p^{2}+13p+42=0$ II. $q^{2}=36$
a) Give answer (A) if p is greater than q.
b) Give answer (B) if p is smaller than q.
c) Give answer (C) if p is equal to q.
d) Give answer (D) if p is either equal to or greater than q.
e) Give answer (E) if p is either equal to or smaller than q.
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 6: I. $2x^{2}+19x+45=0$
II. $2y^{2}+11y+12=0$
a) x = y
b) x> y
c) x < y
d) relationship between xand y cannot be determined
e) x + y
Download SBI Clerk Previous Papers PDF
Question 7: I. $3x^{2}-13x+12=0$
II. $2y^{2}-15y+28=0$
a) x> y
b) x= y
c) x < y
d) relationship between x and y cannot be determined
e) x≤ y
Question 8: I. $x^{2}=16$
II. $2y^{2}-17y+36=0$
a) x > y
b) x > y
c) x < y
d) relationship between x and y cannot be determined
e) $x \leq y$
Question 9: I. $6x^{2}+19x+15=0$
II. $3y^{2}+11y+10=0$
a) x = y
b) x > y
c) x < y
d) $x \geq y$
e) $x \leq y$
Question 10: I. $2x^{2}-11x+15=0$
II. $2y^{2}-11y+14=0$
a) x > y
b) x> y
c) x < y
d) relationship between x and y cannot be determined
e) x ≤ y
Instructions
In the following questions two equations numbered I and
II are given. You have to solve both the equations and
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.
105 SBI Clerk mocks for Rs. 199. Enroll here
15000 Free Banking Solved Questions
Question 11: I. $x^{2}+x-12=0$
II. $y^{2}+2y-8=0$
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 12: I. $4x^{2}-13x+9=0$
II. $3y^{2}-14y+16=0$
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 13: I. $8x^{2}+18x+9=0$
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 the relationship cannot be established.
Question 14: I. $3x^{2}+16x+21=0$
II. $6y^{2}+17y+12=0$
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 15: I. $x^{2}=49$
II. $y^{2}-4y-21=0$
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.
Download SBI Clerk Previous Papers PDF
Answers & Solutions:
1) Answer (B)
$6p^2+5p+1 = 0$
$(2p+1)(3p+1) = 0$
$p = -\frac{1}{2}, -\frac{1}{3}$
$20q^2+9q+1 = 0$
$(4q+1)(5q+1) = 0$
$q = -\frac{1}{4}, -\frac{1}{5}$
$p < q$
2) Answer (A)
$3p^2+2p-1 = 0$
$(3p-1)(p+1) = 0$
$p = -1, \frac{1}{3}$
$2q^2+7q+6 = 0$
$(2q+3)(q+2) = 0$
$q = -2, -\frac{3}{2}$
p > q
3) Answer (D)
$3p^2+15p+18 = 0$
$p^2+5p+6 = 0$
$(p+2)(p+3) = 0$
$p = -3, -2$
$q^2+7q+12 = 0$
$(q+4)(q+3) = 0$
$q = -4, -3$
$p\geq q$
4) Answer (C)
$p = \frac{\sqrt{4}}{\sqrt{9}}$
$p = \frac{2}{3}$
$9q^2-12q+4 = 0$
${(3q-2)}^2 = 0$
$q = \frac{2}{3}$
p = q
5) Answer (E)
$p^2+13p+42 = 0$
$(p+6)(p+7) = 0$
$p = -6, -7$
$q^2 = 36$
$q = -6, 6$
$p\leq q$
6) Answer (C)
$2x^2+19x+45 = 0$
$(2x+9)(x+5) = 0$
$x = -5, -\frac{9}{2}$
$2y^2+11y+12 = 0$
$(2y+3)(y+4) = 0$
$y = -4, -\frac{3}{2}$
x < y
7) Answer (C)
$3x^2-13x+12 = 0$
$(3x-4)(x-3) = 0$
$x = \frac{4}{3}, 3$
$2y^2-15y+28 = 0$
$(2y-7)(y-4) = 0$
$y = \frac{7}{2}, 4$
x < y
8) Answer (E)
$x^2 = 16$
$x = 4, -4$
$2y^2-17y+36 = 0$
$(2y-9)(y-4) = 0$
$y = \frac{9}{2}, 4$
$x \leq y$
9) Answer (D)
$6x^2+19x+15 = 0$
$(3x+5)(2x+3) = 0$
$x = -\frac{5}{3}, -\frac{3}{2}$
$3y^2+11y+10 = 0$
$(3y+5)(y+2) = 0$
$y = -\frac{5}{3}, -2$
$x\geq y$
10) Answer (D)
$2x^2-11x+15 = 0$
$(2x-5)(x-3) = 0$
$x = 3, \frac{5}{2}$
$2y^2-11y+14 = 0$
$(2y-7)(y-2) = 0$
$y = 2, \frac{7}{2}$
relationship between x and y cannot be established
11) Answer (E)
$x^2+x-12 = 0$
$(x-3)(x+4) = 0$
$x = -4, 3$
$y^2+2y-8 = 0$
$(y-2)(y+4) = 0$
$y = -4, 2$
Hence, a relationship can not be established between $x$ and $y$
12) Answer (E)
$4x^2-13x+9 = 0$
$(4x-9)(x-1) = 0$
$x = 1, \frac{9}{4}$
$3y^2-14y+16 = 0$
$(3y-8)(y-2) = 0$
$y = 2, \frac{8}{3}$
relationship between x and y cannot be established
13) Answer (A)
$8x^2+18x+9 = 0$
$(4x+3)(2x+3) = 0$
$x = -\frac{3}{2}, -\frac{3}{4}$
$4y^2+19y+21 = 0$
$(4y+7)(y+3) = 0$
$y = -3, -\frac{7}{4}$
x > y
14) Answer (C)
$3x^2+16x+21 = 0$
$(3x+7)(x+3) = 0$
$x = -3, -\frac{7}{3}$
$6y^2+17y+12 = 0$
$(3y+4)(2y+3) = 0$
$y = -\frac{4}{3}, -\frac{3}{2}$
x < y
15) Answer (E)
$x^2 = 49$
$x = -7, 7$
$y^2-4y-21 = 0$
$(y-7)(y+3) = 0$
$y = -3, 7$
Hence, pairs of (x,y) are (-7,-3), (-7,7), (7,-3) and (7,7). Hence, in some x is less than y and in some x is greater than y. Thus no relation can be established.
