**Quadratic Equation Questions For CMAT Exam**

Download CMAT Quadratic EquationsÂ Questions and Answers PDF covering the important questions. Most expected Quadratic questions with explanations for CMAT 2021 exam.

Download Important Questions On Quadratic Equations For CMAT Exam

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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) $a\leq b$

c) relationship between a & b cannot be established

d) a>b

e) $a\geq b$

**Question 1:Â **I.$2a^{2}+a-1=0$

II.$12b^{2}-17b+6=0$

a)Â $a<b$

b)Â $a\leq b$

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

d)Â $a>b$

e)Â $a\geq b$

**Question 2:Â **I.$a^{2}-5a+6=0$

II. $2b^{2}-13b+21=0$

a)Â $a<b$

b)Â $a\leq b$

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

d)Â $a>b$

e)Â $a\geq b$

**Question 3: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 $a \geq b$

**Question 4:Â **I.$16a^{2}=1$

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

a)Â $a<b$

b)Â $a\leq b$

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

d)Â $a>b$

e)Â $a\geq b$

**Question 5:Â **I.$a^{2}+2a+1=0$

II.$b^{2}=\pm4$

a)Â $a<b$

b)Â $a \leq b$

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

d)Â $a>b$

e)Â $a \geq b$

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**Instructions**

In each of the following question two equations are given you have to solve them and

Give answer (a)if p<q

</q

Give answer (b)if p>q

Give answer (c)if $p\leq$q

Give answer(d)if $p\geq$q

Give answer (e)if p=q

**Question 6:Â **I.$p^{2}-7p=-12$

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

a)Â if p<q

b)Â if p>q

c)Â if $p\leq$q

d)Â if $p\geq$q

e)Â if p=q

**Question 7:Â **I. $12p^{2}-7p=-1$

II. $6q^{2}-7q+2=0$

a)Â if $p < q$

b)Â if $p > q$

c)Â if $p\leq q$

d)Â if $p\geq q$

e)Â if $p = q$

**Question 8:Â **I.$p^{2}+12p+35=0$

II.$2q^{2}+22q+56=0$

a)Â if p < q

b)Â if p>q

c)Â if $p \leq q$

d)Â if $p\geq q$

e)Â if p=q or no relationship can be established

**Question 9:Â **I.$p^{2}-8p+15=0$

II.$q^{2}-5q=-6$

a)Â if p < q

b)Â if p>q

c)Â if $p\leq q$

d)Â if $p\geq q$

e)Â if p=q

**Question 10:Â **I.$2p^{2}+20p+50=0$

II.$q^{2}=25$

a)Â if p<q

b)Â if p>q

c)Â if $p \leq q$

d)Â if $p\geq q $

e)Â if p = q

**Instructions**

For the two given equations I and II—-

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**Question 11: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.

**Question 12:Â **I. $3p^{2}+2p-1=0$ II. $2q^{2}+7q+6=0$

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.

**Question 13:Â **I. $p=\frac{\sqrt{4}}{\sqrt{9}}$ II. $9q^{2}-12q+4=0$

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.

**Question 14:Â **I. $p^{2}+13p+42=0$ II. $q^{2}=36$

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 these questions, two equations numbered I and II are given. You have to solve both the equations and select the appropriate option.

**Question 15:Â **I. $2x^{2}+19x+45=0$

II. $2y^{2}+11y+12=0$

a)Â x = y

b)Â x> y

c)Â x < y

d)Â relationship between xand y cannot be determined

e)Â x + y

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**Question 16:Â **I. $3x^{2}-13x+12=0$

II. $2y^{2}-15y+28=0$

a)Â x> y

b)Â x= y

c)Â x < y

d)Â relationship between x and y cannot be determined

e)Â xâ‰¤ y

**Question 17:Â **I. $x^{2}=16$

II. $2y^{2}-17y+36=0$

a)Â $x > y$

b)Â $x\geq y$

c)Â $x < y$

d)Â relationship between x and y cannot be determined

e)Â $x \leq y$

**Question 18:Â **I. $6x^{2}+19x+15=0$

II. $3y^{2}+11y+10=0$

a)Â x = y

b)Â x > y

c)Â x < y

d)Â $x \geq y$

e)Â $x \leq y$

**Question 19:Â **I. $2x^{2}-11x+15=0$

II. $2y^{2}-11y+14=0$

a)Â x > y

b)Â x> y

c)Â x < y

d)Â relationship between x and y cannot be determined

e)Â x â‰¤ 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 â‰¥ y

c: if x < y

d: if x â‰¤ y

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

**Question 20:Â **I. $x^{2}+x-12=0$

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

a)Â if x > y

b)Â if x â‰¥ y

c)Â if x < y

d)Â if x â‰¤ y

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

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**Answers & Solutions:**

**1)Â AnswerÂ (A)**

$2a^{2}+a-1=0$

We get the factor as:

a=-1, a=0.5

$12b^{2}-17b+6=0$

Solving, we get the factor as,

a= .66, b= .75

Hence, b>a

Option A is the correct option.

**2)Â AnswerÂ (B)**

Soving the quadratic equations we get,

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

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

i.e a=2, a=3

$2b^{2}-13b+21=0$

i.e (b-3.5)(b-3)=0

i.e b= 3.5 and b=3

Hence, we can deduce that $a\leq b$

Therefore, option B is correct.

**3)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.

**4)Â AnswerÂ (D)**

$16a^{2}=1$

Solving we get, a=-.25, a=+.25

$3b^{2}+7b+2=0$

Solving we get, b= -2. b = -1/3

Hence, a>b. Option D is correct.

**5)Â AnswerÂ (C)**

We can easily solve equation I to get a = -1

But we cannot solve $b^{2}=\pm4$. Square root of negative number is not a real number.

Hence, we cannot find a value of b. Therefore, we cannot establish a relationship between a and b.

**6)Â AnswerÂ (B)**

$p^2-7p+12 = 0$

$(p-3)(p-4) = 0$

$p = 3, 4$

$q^2-3q+2 = 0$

$(q-1)(q-2) = 0$

$q = 1, 2$

$\therefore p > q$

**7)Â AnswerÂ (A)**

$12p^2-7p+1 = 0$

$(4p-1)(3p-1) = 0$

$p = \frac{1}{3}, \frac{1}{4}$

$6q^2-7q+2 = 0$

$(2q-1)(3q-2) = 0$

$q = \frac{1}{2}, \frac{2}{3}$

$\therefore p < q$

**8)Â AnswerÂ (E)**

$p^2+12p+35 = 0$

$(p+5)(p+7) = 0$

$p = -5, -7$

$2q^2+22q+56 = 0$

$q^2+11q+28 = 0$

$(q+4)(q+7) = 0$

$q = -4, -7$

As we can see $p$ can be greater than, less than or equal to $q$. No relationship can be established between $p$ and $q$ and hence, option E is the right answer.

**9)Â AnswerÂ (D)**

$p^2-8p+15 = 0$

$(p-3)(p-5) = 0$

$p = 3, 5$

$q^2-5q+6 = 0$

$(q-2)(q-3) = 0$

$q = 2, 3$

$p\geq q$

**10)Â AnswerÂ (C)**

$2p^2+20p+50 = 0$

$p^2+10p+25 = 0$

$(p+5)^2 = 0$

$p = -5$

$q^2 = 25$

$q = 5, -5$

$p\leq q$

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**11)Â AnswerÂ (B)**

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

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

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

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

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

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

$p < q$

**12)Â AnswerÂ (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

**13)Â AnswerÂ (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

**14)Â AnswerÂ (E)**

$p^2+13p+42 = 0$

$(p+6)(p+7) = 0$

$p = -6, -7$

$q^2 = 36$

$q = -6, 6$

$p\leq q$

**15)Â AnswerÂ (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

**16)Â AnswerÂ (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

**17)Â AnswerÂ (E)**

$x^2 = 16$

$x = 4, -4$

$2y^2-17y+36 = 0$

$(2y-9)(y-4) = 0$

$y = \frac{9}{2}, 4$

$x \leq y$

**18)Â AnswerÂ (D)**

$6x^2+19x+15 = 0$

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

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

$3y^2+11y+10 = 0$

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

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

$x\geq y$

**19)Â AnswerÂ (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

**20)Â AnswerÂ (E)**

$x^2+x-12 = 0$

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

$x = -4, 3$

$y^2+2y-8 = 0$

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

$y = -4, 2$

Comparing x and y,

-4 = -4

-4 < 2

3 > -4

3 > 2

Hence, relationship can not be established between $x$ and $y$