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Quant Mania: Quadratic Inequality

Published on Tuesday, January 19, 2016
Dear Readers,
Today we are presenting you Quant Mania on Quadratic Inequality which is very important for your upcoming LIC AAO, Syndicate Bank PO, SIDBI Officers Scale Grade A and other exams. You may expect same questions in your upcoming exams. Try to solve it.

Quant Mania: 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 – 2x - 15 = 0, II. y2 + 5y + 6 = 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. x2 – x – 12 = 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.

3. I. x - √169 = 0, II. y– 169 = 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.

4. I. x2 - 32 =112, II. y - √256 = 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. x2 – 25 = 0, II. y– 9y + 20 = 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.

Answer

1. b;
I. x2  – 2x - 15 = 0
or, x2  – 5x + 2x - 15 = 0
or. x(x - 5) + 3(x – 5) = 0
or, (x – 5) (x + 3) = 0
x = 5, -3

II. y + 5y + 6 = 0
or, y + 3y + 2y + 6 = 0
or, y(y + 3) + 2(y +3) = 0
or, (y + 3) (y +2) = 0
y = -3, -2

Hence, x ≥ y

2. e;
I. x2 – x – 12 = 0
or, x2 – 4x + 3x – 12 = 0
or, x(x – 4) + 3(x - 4) = 0
or, (x – 4) (x + 3) = 0
x = 4, -3

II. y - 3y + 2 = 0
or, y - 2y - y + 2 = 0
or, y(y - 2) - 1(y - 2) = 0
or, (y - 2) (y - 1) = 0
y = 1, 2

Hence, no relation can be established.

3. b;
I. x - √169 = 0
or, x = √169
or, x = 13
II. y2 – 169 = 0
or, y2 – 169 = 0
or, y2 = 169
or, y = ± 13
Hence, x ≥ y

4. c;
I. x2 - 32 =112
or, x2 = 112 + 32
or, x2 = 144
or, x = ± 12

II. y - √256 = 0
or, y = √256
or, y = 16

Hence, x < y

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


II. y – 9y + 20 = 0
or, y – 5y – 4y + 20 = 0
or, y(y – 5)– 4(y - 5) = 0
or, (y – 5) (y – 4) = 0
or, y = 5, 4


Hence, no relation can be established.


Thanks.

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