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# Quadratic Equation Questions For CMAT Exam

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
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 $a \geq b$

Instructions

In each of the following question two equations are given you have to solve them and
</q
Give answer (c)if $p\leq$q
Give answer(d)if $p\geq$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—-

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

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.

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

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.

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

$16a^{2}=1$
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.

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.

$p^2-7p+12 = 0$
$(p-3)(p-4) = 0$
$p = 3, 4$

$q^2-3q+2 = 0$
$(q-1)(q-2) = 0$
$q = 1, 2$

$\therefore p > q$

$12p^2-7p+1 = 0$
$(4p-1)(3p-1) = 0$
$p = \frac{1}{3}, \frac{1}{4}$

$6q^2-7q+2 = 0$
$(2q-1)(3q-2) = 0$
$q = \frac{1}{2}, \frac{2}{3}$

$\therefore p < q$

$p^2+12p+35 = 0$
$(p+5)(p+7) = 0$
$p = -5, -7$

$2q^2+22q+56 = 0$
$q^2+11q+28 = 0$
$(q+4)(q+7) = 0$
$q = -4, -7$

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.

$p^2-8p+15 = 0$
$(p-3)(p-5) = 0$
$p = 3, 5$

$q^2-5q+6 = 0$
$(q-2)(q-3) = 0$
$q = 2, 3$

$p\geq q$

$2p^2+20p+50 = 0$
$p^2+10p+25 = 0$
$(p+5)^2 = 0$
$p = -5$

$q^2 = 25$
$q = 5, -5$

$p\leq q$

$6p^2+5p+1 = 0$
$(2p+1)(3p+1) = 0$
$p = -\frac{1}{2}, -\frac{1}{3}$

$20q^2+9q+1 = 0$
$(4q+1)(5q+1) = 0$
$q = -\frac{1}{4}, -\frac{1}{5}$

$p < q$

$3p^2+2p-1 = 0$
$(3p-1)(p+1) = 0$
$p = -1, \frac{1}{3}$

$2q^2+7q+6 = 0$
$(2q+3)(q+2) = 0$
$q = -2, -\frac{3}{2}$

p > q

$p = \frac{\sqrt{4}}{\sqrt{9}}$
$p = \frac{2}{3}$

$9q^2-12q+4 = 0$
${(3q-2)}^2 = 0$
$q = \frac{2}{3}$

p = q

$p^2+13p+42 = 0$
$(p+6)(p+7) = 0$
$p = -6, -7$

$q^2 = 36$
$q = -6, 6$

$p\leq q$

$2x^2+19x+45 = 0$
$(2x+9)(x+5) = 0$
$x = -5, -\frac{9}{2}$

$2y^2+11y+12 = 0$
$(2y+3)(y+4) = 0$
$y = -4, -\frac{3}{2}$

x < y

$3x^2-13x+12 = 0$
$(3x-4)(x-3) = 0$
$x = \frac{4}{3}, 3$

$2y^2-15y+28 = 0$
$(2y-7)(y-4) = 0$
$y = \frac{7}{2}, 4$

x < y

$x^2 = 16$
$x = 4, -4$

$2y^2-17y+36 = 0$
$(2y-9)(y-4) = 0$
$y = \frac{9}{2}, 4$

$x \leq y$

$6x^2+19x+15 = 0$
$(3x+5)(2x+3) = 0$
$x = -\frac{5}{3}, -\frac{3}{2}$

$3y^2+11y+10 = 0$
$(3y+5)(y+2) = 0$
$y = -\frac{5}{3}, -2$

$x\geq y$

$2x^2-11x+15 = 0$
$(2x-5)(x-3) = 0$
$x = 3, \frac{5}{2}$

$2y^2-11y+14 = 0$
$(2y-7)(y-2) = 0$
$y = 2, \frac{7}{2}$

relationship between x and y cannot be established

$x^2+x-12 = 0$
$(x-3)(x+4) = 0$
$x = -4, 3$

$y^2+2y-8 = 0$
$(y-2)(y+4) = 0$
$y = -4, 2$

Comparing x and y,
-4 = -4
-4 < 2
3 > -4
3 > 2
Hence, relationship can not be established between $x$ and $y$