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Quadratic Equation Practice set for 4 IBPS Clerk 2017

Quadratic Equation Practice set for  4 IBPS Clerk 2017









1 .
Directions (Q. 1-5): Two equations (I) and (II) are given in each question. On the basis of these equations you have to decide the relation between ‘x’ and ‘y’ and give answer.

(1) if x > y 
(2) if x < y 
(3) if x  y
(4) if x  y 
(5) if x = y or no relation can be established between ‘x’ and ‘y’.

Q. I. 6x219x+15=0
II.10y229y+21=0
A). x > y
B). x < y
C).  y
D).  y
Answer & Explanation
Answer : Option C
 
Explanation :
 
6x219x+15=0
3x(2x3)5(2x3)=0 
(3x5)(2x3)=0 
x = 53 , 32

10y215y14y+21=0
5y(2y3)7(2y3)=0
(5y7)(2y3)=0
y = 75 , 32
xy
Hide Answer

2 .I. 12 x2 + 11x - 56 = 0 
II. 4 y2 - 15y + 14 = 0
A). x > y
B). x < y
C).  y
D).  y
Answer & Explanation
Answer : Option D
 
Explanation :
 
I. 12x2+32x21x56 = 0
or, 4x(3x + 8) - 7(3x + 8) = 0
or, (4x - 7) (3x + 8) = 0
X = 74 , 83

II. 4y28y7y+14= 0
or, 4y(y - 2) - 7(y - 2) = 0
or, (4y - 7) (y - 2) = 0
y = 2, 74

xy
Hide Answer

3 .I. 3x2+13x+12= 0
II.y2+9y+20= 0
A). x > y
B). x < y
C).  y
D).  y
Answer & Explanation
Answer : Option A
 
Explanation :
 
I . 3x2 + 9x + 4x + 12 = 0
or, 3x(x + 3) + 4(x + 3) = 0
or, (3x + 4) (x + 3) = 0
x = 43, -3

II.y2 + 5y + 4y + 20 = 0
or, y(y + 5) + 4(y + 5) = 0
or, (y + 4) (y + 5) = 0
y = - 4, - 5

x>y
Hide Answer

4 .I. 8x2 - 15x + 7 = 0 
II. 2y2 - 7y + 6 = 0
A). x > y
B). x < y
C).  y
D).  y
Answer & Explanation
Answer : Option B
 
Explanation :
 
I. 8x2 - 8x - 7x + 7 = 0
or, 8x(x - 1) -7(x - 1) = 0
or, (8x - 7) (x - 1) = 0
x = 78 , 1

II. 2y2 - 4y - 3y + 6 = 0
or, 2y(y - 2) -3(y - 2) = 0
or, (y - 2) (2y - 3) = 0
y = 2 , 32

x<y
Hide Answer

5 .I. 7x - 3y = 13 
II. 5x + 4y = 40
A). x > y
B). x < y
C).  y
D).  y
Answer & Explanation
Answer : Option B
 
Explanation :
 
Eqn (I) × 4 + Eqn (II) × 3 

28x - 12y = 52
15x + 12y = 120
-----------------------
43x  = 172
-----------------------

x = 4 & y = 5 

x<y
Hide Answer

6 .I.2x2 - 11x + 15 = 0 
II.21y2 - 23y + 6 = 0
A). x > y
B). x < y
C).  y
D). x = y or no relation can be established between ‘x’ and ‘y’.
Answer & Explanation
Answer : Option A
 
Explanation :
 
I. 2x2 - 6x - 5x + 15 = 0
or,2x(x - 3) - 5(x - 3) = 0
or, (2x - 5) (x - 3) = 0
x = 3 , 52

II. 21y2 - 14y - 9y + 6 = 0
or,7y(3y - 2) - 3 (3y - 2) = 0
or,(7y - 3)(3y - 2) = 0
y = 37 , 23

x>y
Hide Answer

7 .I. 5x2 - 16x + 11= 0 
II. 5y2 - 3y - 2 = 0
A). x>y
B).  y
C). x< td>
D).  y
Answer & Explanation
Answer : Option B
 
Explanation :
 
I. 5x2 - 5x - 11x + 11 = 0
or,5x(x - 1) - 11(x - 1) = 0
or,(x - 1) (5x - 11) = 0

x = 1 , 115

II. 5y2 - 5y + 2y - 2 = 0
or, 5y (y - 1) + 2(y - 1) = 0
or, (5y + 2)(y - 1) = 0

y = 1 , - 25

 y
Hide Answer

8 .I. x2 + 11x + 28 = 0 
II. 2y2 + 13y + 20 = 0
A). x > y
B). x < y
C). xy
D). xy
Answer & Explanation
Answer : Option D
 
Explanation :
 
I. x2 + 4x + 7x + 28 = 0
or, x(x + 4) +7(x + 7) = 0
or, (x + 4) (x + 7) = 0

x = - 4, - 7

II. 2y2 + 8y + 5y + 20 = 0
or, 2y(y + 4) + 5(y + 4) = 0
or, (y + 4) (2y + 5) = 0
y = -4, -52

 y
Hide Answer

9 .I. 6x2 + 29x + 35 = 0
II. 3y2 + 19y + 30 = 0
A). x>y
B). x<y
C). xy
D). xy
Answer & Explanation
Answer : Option A
 
Explanation :
 
I. 6x2 + 15x + 14x + 35 = 0
or, 3x(2x + 5) + 7(2x + 5) = 0
or,(3x + 7) (2x + 5) = 0
x = - 73 , - 52

II. 3y2 + 9y + 10y + 30 = 0
or, 3y(y + 3) +10(y + 3) = 0
or,(3y + 10) (y + 3) = 0
y = 3, -103

x>y
Hide Answer

10 .I. 2x + 5y = 6 
II. 5x + 11y = 9
A). x>y
B). x<y
C). xy
D). xy
Answer & Explanation
Answer : Option B
 
Explanation :
 
eq (I) × 5 - eq (II) × 2

10x + 25y = 30
10x ± 22y = 18
 - 
---------------------
 3y = 12
y = 4 and x = - 7 

x<y 

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