## Sunday, 23 April 2017

Directions (Q. 1-5): In each question, two equations numbered I and II are given. You have to solve both the equations and mark an appropriate answer.
a)   x < y
b)   x ≥ y
c)   x > y
d)   x ≤ y
e)   Relation between x and y can't be established

1). I. 9x2 - 29x + 22 = 0
II. 6y2 – 31y + 39 = 0

2). I. x2 = 11449
II. y = √11449

3). I. 6x2 - 25x – 14 = 0
II. 8y2 – 35y + 12 = 0

4). I. 5x – 4y = 83
II. 6x + 3y = 45

5). I. 9x2 + 6√7 x + 7 = 0
II. 4y2 + 4√3 y + 3 = 0

Directions (Q. 6-10): In each question two equations numbered I and II are given. You have to solve both the equations and mark the appropriate answer.
6). I. 14x2+27x+9=0
II. 2y2+ 9y +10 = 0
a)   x ≥ y
b)   x<y
c)   x = y or no relation can be established
d)   x ≤ y
e)   x > y

7). I. 20x2— 53x + 18=0
II. 5y2— 37y + 32 = 0
a)   x ≤ y
b)   x>y
c)   x≥y
d)   x<y
e)   x = y or no relation can be established

8). I.12x2— 25x+ 13 = 0
II. 5y2— 12y + 4 = 0
a)   x<y
b)   x≥y
c)   x≤y
d)   x = y or no relation can be established
e)   x>y

9). I. 2x2-7x-22=0
II.6y2-13y-19 = 0
a)   x>y
b)   x = y or no relation can be established
c)   x≤y
d)   x≥y
e)   x<y

10). I. 10x2 - (15 + 8√3)x + 12√3 = 0
II. 6y2 + (8 + 3√3)y + 4√3 = 0
a)   x≤y
b)   x≥y
c)   x<y
d)   x>y
e)   x = y or no relation can be established

Explanation:

1). A) I. 9x2 - 29x + 22 = 0
x = (-18/9), (-11/9)
x = 2, 11/9
II. 6y2 – 31y + 39 = 0
y = (-18/6), (-13/6)
y = 3, 13/6
hence x < y

2). D) I. x2 = 11449
x = √11449 = ±107
II. y = √11449 = 107
Hence x ≤ y

3). E) I. 6x2 - 25x – 14 = 0
x = (3/6), (-28/6)
x = -1/2, 14/3
II. 8y2 – 35y + 12 = 0
y = (-3/8), (-32/8)
y = 3/8, 4
hence no relation can be established.

4). C) I. 5x – 4y = 83     ..(i)
II. 6x + 3y = 45      …(ii)
Solving equation (i)×3 + (ii)×4, we get
(15x – 12y = 249) + (24x + 12y = 180)
x = (429/39) = 11
putting the value of x in equation (i), we get
55 – 4y = 83
-4y = 83-55
y = -28/4 = -7
hence x > y

5). A) I. 9x2 + 6√7 x + 7 = 0
Or, (3x + √7)2 = 0
Or, 3x = -√7
x = -3/√7
II. 4y2 + 4√3 y + 3 = 0
Or, (2y - √3)2 = 0
Or, 2y = √3
y = √3/2
hence x < y

6). E) I. 14x2+27x+9=0
x = (21/4), (6/14)
x = -3/2, -3/7
II. 2y2+ 9y +10 = 0
y = (5/2), (4/2)
y = -5/2, -2
hence x > y

7). E) I. 20x2— 53x + 18=0
x = (-45/20), (-8/20)
x = 9/4, 2/5
II. 5y2— 37y + 32 = 0
y = (-32/5), (-5/5)
y = 6.4, 1
Hence no relation can be established.

8). B) I.12x2— 25x+ 13 = 0
x = (-13/12), (-12/12)
x = 13/12, 1
II. 5y2— 12y + 4 = 0
y = (-10/5), (-2/5)
y = 2, 2/5
Hence relationship can’t be established.

9). B) I. 2x2-7x-22=0
x = (-11/2), (4/2)
x = 5.5, -2
II.6y2-13y-19 = 0
y = (-19/6), (6/6)
y = 19/6, -1
Hence relation can’t be established.

10). D) I. 10x2 - (15 + 8√3)x + 12√3 = 0
x = (-15/10), (-8√3 /10)
x = 3/2, 4√3/5
II. 6y2 + (8 + 3√3)y + 4√3 = 0
y = (-4/3), (-√3 /2)
hence x > y