Quant Mania: Quadratic Inequality

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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