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Quadratic Equation Practice set for 2 IBPS Clerk 2017

Quadratic Equation Practice set 4 for IBPS Clerk 2017


1 .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) if x>y
(2) if xy
(3) if x<y
(4) if xy
(5) if x = y or relationship between x and y cannot be established

Q. 
I.11x + 5y = 117 
II. 7x + 13y = 153
A). x>y
B). xy
C). x<y
D). xy
Answer & Explanation
Answer : Option C
Explanation :
eqn (I) × 7
eqn (II) × 11 
77x + 35y = 819
- 77x ± 143y = 1683
------------------------------
- 108y = - 864
y = 8, x = 7 

ie x < y
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2 .I.6x2 + 51x + 105 = 0 
II. 2y2 + 25y + 78 = 0
A). x>y
B). xy
C). x<y
D). xy
Answer & Explanation
Answer : Option C
Explanation :
I. 6x2 + 21x + 30x + 105 = 0
or, 3x(2x + 7) + 15(2x + 7) = 0
or, (3x + 15) (2x + 7) = 0
x = -5 , -72

II. 2y2 + 12y + 13y + 78 = 0
or, 2y(y + 6) + 13(y + 6) = 0
or, (2y + 13) (y + 6) = 0
y = -132 , -6

x<y
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3 .I.6x + 7y = 52
II. 14x + 4y = 35
A). x>y
B). xy
C). x<y
D). xy
Answer & Explanation
Answer : Option C
Explanation :
eqn (I) × 4
eqn (II) × 7

24x + 28y = 208
98x ± 28y = 245
-
----------------------
- 74x = - 37
x = 12, y = 7

x<y
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4 .I.x2 + 11x + 30 = 0 
II. y2 + 12y + 36 = 0
A). x>y
B). xy
C). x<y
D). xy
Answer & Explanation
Answer : Option B
Explanation :
I. x2 + 5x + 6x + 30 = 0
or, x(x + 5) + 6(x + 5) = 0
or, (x + 5) (x + 6) = 0
x = - 5, - 6

II. y2 + 12y + 36 = 0
or, (y+6)2 = 0
or, y + 6 = 0
y = - 6

ie x  y
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5 .I.2x2 + x - 1 = 0 
II. 2y2 - 3y + l = 0
A). x>y
B). xy
C). x<y
D). xy
Answer & Explanation
Answer : Option D
Explanation :
I. 2x2 + 2x - x - 1 = 0
or, 2x(x + 1) - 1(x + 1) = 0
or, (2x - 1) (x + 1) = 0
x = 12 , -1

II. 2y2 - 2y - y + 1 = 0
or, 2y(y - 1) - 1(y - 1) = 0
or, (2y - 1)(y - 1) = 0
y = 12, 1

i.e., xy
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6 .Directions (Q.6-10) : In the following questions three equations numbered I, II and III are given. You have to solve all the equations either together or separately, or two together and one separately, or by any other method and give answer If
(1) x < y = z 
(2) x < y < z 
(3) x < y > z 
(4) x = y > z
(5) x = y = z or if none of the above relationship is established

Q.
I. 7x + 6y + 4z = 122 
II. 4x + 5y + 3z = 88 
III. 9x + 2y + z = 78
A). x < y = z
B). x < y < z
C). x < y > z
D). x = y > z
Answer & Explanation
Answer : Option A
Explanation :
7x + 6y + 4z = 122 ... (i)
4x + 5y + 3z = 88 ... (ii)
9x + 2y + z = 78 ... (iii)
From (i) and (ii)
5x - 2y = 14... (iv)
From (ii) and (iii)
23x + y = 146 ... (v)
From (iv) and (v),
x = 6, y = 8
Putting the value of x and y in eqn (i), we get
z = 8

:. x < y = z
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7 .I. 7x + 6y =110
II. 4x + 3y = 59
III. x + z = 15
A). x < y = z
B). x < y < z
C). x < y > z
D). x = y > z
Answer & Explanation
Answer : Option C
Explanation :
7x + 6y = 110 ... (i)
4x + 3y = 59 ... (ii)
x + z = 15 ... (iii)
From eqn (i) and (ii), x = 8, y = 9
Put the value of x in eqn (iii).
Then, z = 7

x < y > z
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8 .I. x = [(36)12×[1296]14]
II. 2y + 3z = 33 
III. 6y + 5z = 71
A). x < y = z
B). x < y < z
C). x < y > z
D). x = y < z
Answer & Explanation
Answer : Option D
Explanation :
x = (62)12×(64)14
6×6 = 6 ..(i)
2y + 3z = 33 ... (ii)
6y + 5z = 71 ... (iii)
From eqn (ii) and (iii),
y = 6 and z = 7

x = y < z
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9 .I. 8x + 7y= 135 
II. 5x + 6y = 99
III. 9y + 8z = 121
A). x < y = z
B). x < y < z
C). x < y > z
D). x = y > z
Answer & Explanation
Answer : Option D
Explanation :
8x + 7y = 135 ... (i)
5x + 6y = 99 ... (ii)
9y + 8z = 121 ... (iii)
From eqn (i) and (ii),
x = 9, and y = 9
Putting the value of y in eqn (iii),
z = 5

:. x = y > z
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10 .I. (x+y)3 = 1331 
II. x - y + z = 0
III. xy = 28
A). x < y = z
B). x < y < z
C). x < y > z
D). x = y = z or if none of the above relationship is established
Answer & Explanation
Answer : Option D
Explanation :
(x+y)3 = 1331
or, x + y = 11 ... (i)
(x+y)2 = 121
(xy)2 + 4xy = 121
x - y = 3... (ii)
[value of xy from eqn (iii)]
From eqn (i) and (ii), x = 7, y = 4
Put the value x and y in the eqn
x - y + z = 0
7 - y + z = 0
3 + z = 0
z = -3
Hide Answer

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