**Quadratic Equations Questions for SBI Clerk Set-2**

Download SBI Clerk Quadratic EquationsÂ Questions & Answers Set-2 PDF for SBI Clerk Prelims and Mains exam. Very Important SBI Clerk Quadratic Equations Questions with solutions.

Download Quadratic Equations Questions Set-2

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**Instructions**

In each of these question two equations I & II with variables a & b are given You have to solve both the equations to find the values of a & b

Mark answer if

a) a<b

</b

<b

</b

<b

b) $a\leq b$

c) relationship between a & b cannot be established

d) a>b

e) $a\geq b$

**Question 1:Â **I.$a^{2}+5a+6=0$

II.$b^{2}+7b+12=0$

a)Â $a<b$

b)Â $a \leq b$

c)Â Relationship between $a$ & $b$ cannot be established

d)Â $a>b$

e)Â $a \geq b$

**Instructions**

For the two given equations I and II—-

**Question 2:Â **I. $6p^{2}+5p+1=0$

II. $20q^{2}+9q=-1$

a)Â Give answer (A) if p is greater than q.

b)Â Give answer (B) if p is smaller than q.

c)Â Give answer (C) if p is equal to q.

d)Â Give answer (D) if p is either equal to or greater than q.

e)Â Give answer (E) if p is either equal to or smaller than q.

**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 â‰Ą y

c: if x < y

d: if x â‰¤ y

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

**Question 3:Â **I. $x^{2}=49$

II. $y^{2}-4y-21=0$

a)Â if x > y

b)Â if x â‰Ą y

c)Â if x < y

d)Â if x â‰¤ y

e)Â if x = y or the relationship 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

**Question 4:Â **I. $4x^2 + 15x + 14 = 0$

II. $6y^2 + 10y = 0$

a)Â If x < y

b)Â If x > y

c)Â If x â‰¤ y

d)Â If x â‰Ą y

e)Â If relationship between x and y cannot be determined

**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 5: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

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**Instructions**

In the following questions three equations numbered I, II and III are given.

You have to solve all the equations either together or separately, or two together and one

separately, or by any other method and Give answer

a: x< y=z

b: x â‰¤ y < z

c: x < y > z

d: x = y > z

e: x = y = z or if none of the above relationship is established

**Question 6:Â **I. 8x+ 7y = 135

II. 5x + 6y = 99

III. 9y + 8z = 121

a)Â x< y=z

b)Â x â‰¤ y < z

c)Â x < y > z

d)Â x = y > z

e)Â x = y = z or if none of the above relationship is established

**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 â‰Ą y

Give answer c: if x < y

Give answer d: if x â‰¤ y

Give answer e: if x = y or the relationship between x and y cannot be established.

**Question 7:Â **I.$3x^{2}-13x-10=0$

II.$3y^{2}+10y-8=0$

a)Â if x >y

b)Â if x â‰Ą y

c)Â if x < y

d)Â if x â‰¤ y

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

**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 â‰Ą y

Give answer c: if x < y

Give answer d: if x â‰¤ y

Give answer e: if x = y or the relationship cannot be established.

**Question 8:Â **I.Â $2x^{2}+21x+10=0$

II.Â $3y^{2}+13y+14=0$

a)Â if x > y

b)Â if x â‰Ą y

c)Â if x < y

d)Â if x â‰¤ y

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

**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 â‰Ą 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 9:Â **Three equal circles are drawn on a triangle ABC, with points A, B and C as the centres. Radius of each of the

circle is equal to half of the side of the triangle ABC. (Figure not to the scale)Area of shaded region 1 = $128\frac{1}{3}cm^{2}$

**Quantity :**

I. The area of the shaded region 2 ( in cm^{2} )

II. 30 cm^{2}

a)Â If quantity I â‰Ą quantity II

b)Â If quantity I > quantity II

c)Â If quantity I < quantity II

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

e)Â If quantity quantity II

**Question 10:Â **There are three positive numbers- a, b and c. The average of a and b is less than the average of b and c by 1.

Quantity :

I. Value of c.

II. Value of a

a)Â If quantity I â‰Ą quantity II

b)Â If quantity I > quantity II

c)Â If quantity I < quantity II

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

e)Â If quantity quantity II

**Answers & Solutions:**

**1)Â AnswerÂ (E)**

$a^{2}+5a+6=0$

i.e (a+2)(a+3)=0

i.e a=-2, a=-3

.$b^{2}+7b+12=0$

i.e (b+4)(b+3)=0

i.e b=-4, b=-3

Hence, we can deduce that $a \geq b$.

Therefore, option E is correct.

**2)Â AnswerÂ (B)**

$6p^2+5p+1 = 0$

$(2p+1)(3p+1) = 0$

$p = -\frac{1}{2}, -\frac{1}{3}$

$20q^2+9q+1 = 0$

$(4q+1)(5q+1) = 0$

$q = -\frac{1}{4}, -\frac{1}{5}$

$p < q$

**3)Â AnswerÂ (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.

**4)Â AnswerÂ (A)**

I : 4$x^{2}$ + 15x + 14 = 0

=> 4$x^{2}$ + 8x + 7x + 14 = 0

=> 4x ( x + 2) + 7 ( x + 2) = 0

=> (4x + 7) (x + 2) = 0

=> x = -7/4 , -2

II : 6$y^{2}$ + 10y = 0

=> y (6y + 10) = 0

=> y = 0 , -5/3

Since, 0 > -7/4 , -2 & -5/3 > -7/4 , -2

=> **y > x**

**5)Â AnswerÂ (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.

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**6)Â AnswerÂ (D)**

Applying the operationÂ : Eqn(II) $\times 8$ – Eqn(I) $\times 5$

=> $(40x – 40x) + (48y – 35y) = (792 – 675)$

=> $13y = 117$

=> $y = \frac{117}{13} = 9$

Putting in Eqn (I), we getÂ :

=> $8x + 63 = 135$

=> $8x = 135 – 63 = 72$

=> $x = \frac{72}{8} = 9$

From Eqn(III)

=> $9 \times 9 + 8z = 121$

=> $8z = 121 – 81 = 40$

=> $z = \frac{40}{8} = 5$

$\therefore x = y > z$

**7)Â AnswerÂ (E)**

I.$3x^{2} – 13x – 10 = 0$

=> $3x^2 – 15x + 2x – 10 = 0$

=> $3x (x – 5) + 2 (x – 5) = 0$

=> $(x – 5) (3x + 2) = 0$

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

II.$3y^{2} + 10y – 8 = 0$

=> $3y^2 + 12y – 2y – 8 = 0$

=> $3y (y + 4) – 2 (y + 4) = 0$

=> $(y + 4) (3y – 2) = 0$

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

$\therefore$ No relation can be established.

**8)Â AnswerÂ (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.$3y^{2} + 13y + 14 = 0$

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

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

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

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

$\therefore$Â No relation can be established.

**9)Â AnswerÂ (A)**

**10)Â AnswerÂ (B)**

Average of a and b is less than the average of b and c by 1

=> $\frac{b + c}{2} – \frac{a + b}{2} = 1$

=> $(b + c) – (a + b) = 2$

=> $c – a = 2$

$\because$ Difference between c and a is positive.

=> $c > a$

$\therefore$ Quantity I > Quantity II

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