New Student Offer - Use Code HELLO

Register Now

Quant Mania: Quadratic Inequality (with Detailed Solution)

Published on Tuesday, February 09, 2016
Dear Readers,
Today we are presenting you Quant Mania on Quadratic Inequality which is very important for your upcoming LIC AAO and other exams. You may expect same questions in your upcoming exams. Try to solve it.

Quadratic Inequality

Directions (Q. 1-5): In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.

1. I. x2 – 25 = 0, II. y2 – 3y - 18 = 0
a) if x > y
b) if x ≥ y
c) if x < y
d) if x ≤ y
e) if x = y or no relation can be established between x and y.
2. I. 2x2 – 3x + 1 = 0, II. 2y2 - 9y + 9 = 0
a) if x > y
b) if x ≥ y
c) if x < y
d) if x ≤ y
e) if x = y or no relation can be established between x and y.

3. I. x2 = 4y, II. y = (729)1/3
a) if x > y
b) if x ≥ y
c) if x < y
d) if x ≤ y
e) if x = y or no relation can be established between x and y.

4. x2 – 5x + 6 = 0, II. y2 - 3y + 2 = 0
a) if x > y
b) if x ≥ y
c) if x < y
d) if x ≤ y
e) if x = y or no relation can be established between x and y.

5. I. 3x2 – 19x + 20=0, II. (3y – 4) (y – 5) = 0
a) if x > y
b) if x ≥ y
c) if x < y
d) if x ≤ y
e) if x = y or no relation can be established between x and y.

Answers with Solution

1. e;
I. x2 – 25 = 0
or, x2 = 25
or, x = ± 5

II. y2 – 3y - 18 = 0
or, y2 - 6y + 3y - 18= 0
or, y(y - 6) + 3(y - 6) = 0
or, (y - 6) (y + 3) = 0
or, y = 6, -3
Hence, no relation can be established.

2. c;
I. 2x2 – 3x + 1 = 0
or, 2x2 – 2x - 1x + 1 = 0
or, 2x(x – 1) - 1(x - 1) = 0
or, (2x – 1) (x - 1) = 0
x = ½, 1

II. 2y2 - 9y + 9 = 0
or, 2y2 - 6y - 3y + 9 = 0
or, 2y(y - 3) - 3(y - 3) = 0
or, (2y - 3) (y - 3) = 0
y = 3/2, 3
Hence, x < y

3. c;
II. y = (729)1/3
y = 9


I. x2 = 4y
or, x2 = 4 * 9 = 36
or, x = ±6
Hence, 
x < y

4. b;
I. x2 – 5x + 6 = 0
or, x2 – 3x - 2x + 6 = 0
or. x(x - 3) - 2(x – 3) = 0
or, (x - 3) (x - 2) = 0
x = 3, 2

II. y2 - 3y + 2 = 0
or, y2 - 2y - y + 2 = 0
or, y(y - 2) - 1(y - 2) = 0
or, (y - 2) (y - 1) = 0
y = 1, 2
Hence, x ≥ y

5. e;
I. 3x2 – 19x + 20 = 0,
or, 3x2 – 15x – 4x+ 20 = 0
or, 3x(x – 5) – 4(x – 5) = 0
or, (3x – 4) (x – 5) = 0
x = 4/3, 5

II. (3y – 4) (y – 5) = 0
y = 4/3, 5
Hence, x = y


Thanks.

Can I help you?

ramandeep singh

Hey I am Ramandeep Singh. I am determined to help students preparing for RBI, SEBI, NABARD and IBPS exams. Do you want me to help you ?

Join my class here
    Follow me:
Close Menu
Close Menu