# Quadratic Equation Questions for IBPS Clerk Set-3 PDF

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## Quadratic Equation Questions for IBPS Clerk Set-3 PDF

Download important  Quadratic Equation Questions set-3 PDF based on previously asked questions in IBPS Clerk and other Banking Exams. Practice Scheduling Questions and Answers 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. $3x^{2} + 23x+ 44 = 0$
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 2: 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 3: 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

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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 :
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 4: I. $15x^{2} + 26x + 8 = 0$
II. $25y^{2} + 15y + 2 = 0$

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

Question 5: I. $10x^{2} + 21x + 8 = 0$
II. $5y^{2} + 19y + 18 = 0$

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 the given questions, two quantities are given, one as Quantity I and another as Quantity II. You have to determine relationship between two quantities and choose the appropriate option.

a: If quantity I ≥ quantity II
b: If quantity I > quantity II
c: If quantity I < quantity II
d: If quantity I = quantity II or the relationship cannot be established from the information that is given
e: If quantity quantity II

Question 6: Ram invested P in scheme A and 2P in scheme B, for two years each. Scheme A offers simple interest p.a. Scheme B offers compound interest (compounded annually) at the rate of 10% p.a. Respective ratio between the interest earned from scheme A and that earned from scheme B was 8 : 21.
Quantity :
I. Rate of interest offered by scheme A.
II. Rate of interest offered by scheme C (simple interest p.a.), when 1,600/- is invested for 3 years earns an interest of 384/-.

a) If quantity I ≥ quantity II

b) If quantity I > quantity II

c) If quantity I < quantity II

d) If quantity I = quantity II or the relationship cannot be established from the information that is given

e) If quantity quantity II

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 7: 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

Instructions

In each of these questions an equation is given with a question mark (?) in place of the correct figure on the right hand side which satisfies the equality. Based on the values on the left hand side and the symbol of equality given, you have to decide which of the following figures will satisfy the equality and thus come in place of the question mark.
Symbols and Stand for

>   greater than

=  equal to

<  lesser than

$\geq$ either greater than or equal to

$\leq$  either lesser than or equal to

Question 8: $\pm [(35 \div 2) – (43.5 \div 3)] \leq ?$

a) – 3

b) 9

c) $\sqrt{9}$

d) 3

e) ± 3

Question 9: $[175-(52 – 72)] \geq ?$

a) +195

b) -195

c) 195

d) 196

e) $\sqrt{195}$

Question 10: $-((\sqrt{225}-\sqrt{144})\times-(1.5))=?$

a) 4.5

b) – 4.5

c) 5.4

d) – 5.4

e) 4.8

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

I. $15x^{2} + 26x + 8 = 0$

=> $15x^2 + 6x + 20x + 8 = 0$

=> $(3x + 4) (5x + 2) = 0$

=> $x = \frac{-4}{3} , \frac{-2}{5}$

II. $25y^{2} + 15y + 2 = 0$

=> $25y^2 + 5y + 10y + 2 = 0$

=> $(5y + 2) (5y + 1) = 0$

=> $y = \frac{-2}{5} , \frac{-1}{5}$

$\therefore x \leq y$

I. $10x^{2} + 21x + 8 = 0$

=> $10x^2 + 5x + 16x + 8 = 0$

=> $(5x + 8) (2x + 1) = 0$

=> $x = \frac{-8}{5} , \frac{-1}{2}$

II. $5y^{2} + 19y + 18 = 0$

=> $5y^2 + 10y + 9y + 18 = 0$

=> $(5y + 9) (y + 2) = 0$

=> $y = \frac{-9}{5} , -2$

$\therefore x > y$

$S.I. = \frac{P \times R \times T}{100}$

$C.I. = P [(1 + \frac{R}{100})^T – 1]$

Rate of interest for scheme B = 10 %

Time period in each scheme = 2 years

Acc to ques,

=> $\frac{\frac{P \times R \times 2}{100}}{2P [(1 + \frac{10}{100})^2 – 1]} = \frac{8}{21}$

=> $\frac{\frac{R}{50}}{\frac{21}{50}} = \frac{8}{21}$

=> $\frac{R}{21} = \frac{8}{21}$

=> $R = 8 \%$

Quantity I = 8 %

Quantity II : $384 = \frac{1600 \times R \times 3}{100}$

=> $R = \frac{384}{48} = 8 \%$

$\therefore$ Quantity I = Quantity II

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

$\pm [(35 \div 2) – (43.5 \div 3)]$
= ± [17.5-14.5]
= ± 3
Since ± 3 $\leq$ 3

$-((\sqrt{225}-\sqrt{144})\times-(1.5))=?$
= – (15-12) $\times$ (-1.5)
= (-3)$\times$(-1.5) =4.5