Hello Aspirants. Welcome to Online Quantitative Aptitude Section in AffairsCloud.com. Here we are creating sample questions in Quadratic Equations which is common for all the competitive exams. We have included Some questions that are repeatedly asked in bank exams !!!
You have to solve equation I and II ,Give answer
1)If X>Y
2)If X<Y
3)If X≥ Y
4)If X≤ Y          Â
5)If X=Y or cannot be established
- I. X2 Â + 3X – 18= 0
II.2Y2 + 14Y + 24 = 05)If X=Y or cannot be established
Explanation :
(x+6)(x-3)=0
x =3,-6
(2y+6)(y+4)=0
Y=-4,-6/2 = -4,-3
3,-3,-4,-6 =>xyyx
No relation - I. 2X2 Â – 9X + 7= 0
II. 6Y2 + 7Y – 10 = 01)If X>Y
Explanation :
(2x-7)(x-1)=0
X=1,7/2
(y+2)(6y-5) = 0
y=5/6,-2
3.5, 1, 0.83, -2 =>xxyy
x>y - I. 6X2 + 11X + 3= 0
II. 3Y2 – 2Y – 1 = 04)If X≤ Y
Explanation :
(2x+3)(3x+1) = 0
X= -1/3, -3/2
(y-1)(3y+1) = 0
Y = 1, -1/3
1, -0.33, -0.33, -1.5 => yyxx
X≤ Y - I. X2 – 12X + 35= 0
II. Y2 + Y – 30 = 03)If X≥ Y
Explanation :
(x-5)(x-7) = 0
X = 7,5
(y-5)(y+6) =0
Y = 5, -6
7, 5, 5, -6 = xxyy
X≥ Y - I. X2 = 9
II. Y2 + 3Y + 2 = 05)If X=Y or cannot be established
Explanation :
X = +3, -3
(y+1)(y+2) = 0
Y = -1, -2
3, -1, -2, -3 => xyyx - I. X2 = 36
II. Y2 – 3Y – 54 = 05)If X=Y or cannot be established
Explanation :
X = +6, -6
(y-9)(y+6) = 0
Y = 9, -6
9,6,-6,-6 = yxyx - I. X2 – 4X + 3= 0
II. Y2 – 8Y + 15 = 04)If X≤ Y
Explanation :
(x-3)(x-1) = 0
X=1,3
(y-5)(y-3) = 0
Y = 5,3
5,3,3,1 = yyxx
X≤ Y - I. 2X2 + 15X + 25= 0
II. 3Y2 + 29Y + 56 = 05)If X=Y or cannot be established
Explanation :
(x+5)(2x+5) = 0
X= -5, -5/2
(y+7)(3y+8) = 0
Y= -7, -8/3
-2.5,-2.6,-5,-7 =>yxxy - I. 3X2 – 19X + 28= 0
II. 4Y2 – 29Y + 45 = 05)If X=Y or cannot be established
Explanation :
(3x-7)(x-4) = 0
X=4, 7/3
(y-5)(4y-9) = 0
Y =5, 9/4
5, 4, 2.3, 2.25 = yxxy - I. X2 Â – 13X + 42= 0
II. Y2 – 9Y + 20 = 01)If X>Y
Explanation :
(x-7)(x-6) = 0
X=7,6
(y-5)(y-4) = 0
Y =5,4
7,6,5,4, => xxyy
X>Y
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