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# Quadratic Equation Practice set for 2 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 $x\ge y$
(3) if $x
(4) if $x\le y$
(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$x>y$ B). x≥y$x\ge y$ C). x
 Answer : Option C Explanation : eqn (I) × 7eqn (II) × 11 $\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}$77x + 35y = 819- 77x ± 143y = 1683------------------------------- 108y = - 864y = 8, x = 7 ie x < y

2 .I.$6{x}^{2}$ + 51x + 105 = 0
II. $2{y}^{2}$ + 25y + 78 = 0
 A). x>y$x>y$ B). x≥y$x\ge y$ C). x
 Answer : Option C Explanation : I. 6x2$6{x}^{2}$ + 21x + 30x + 105 = 0or, 3x(2x + 7) + 15(2x + 7) = 0or, (3x + 15) (2x + 7) = 0x = -5 , -72$\frac{7}{2}$II. 2y2$2{y}^{2}$ + 12y + 13y + 78 = 0or, 2y(y + 6) + 13(y + 6) = 0or, (2y + 13) (y + 6) = 0y = -132$\frac{13}{2}$ , -6x

3 .I.6x + 7y = 52
II. 14x + 4y = 35
 A). x>y$x>y$ B). x≥y$x\ge y$ C). x
 Answer : Option C Explanation : eqn (I) × 4eqn (II) × 724x + 28y = 20898x ± 28y = 245------------------------ 74x = - 37x = 12$\frac{1}{2}$, y = 7x

4 .I.${x}^{2}$ + 11x + 30 = 0
II. ${y}^{2}$ + 12y + 36 = 0
 A). x>y$x>y$ B). x≥y$x\ge y$ C). x
 Answer : Option B Explanation : I. x2${x}^{2}$ + 5x + 6x + 30 = 0or, x(x + 5) + 6(x + 5) = 0or, (x + 5) (x + 6) = 0x = - 5, - 6II. y2${y}^{2}$ + 12y + 36 = 0or, (y+6)2$\left(y+6{\right)}^{2}$ = 0or, y + 6 = 0y = - 6ie x ≥$\ge$ y

5 .I.$2{x}^{2}$ + x - 1 = 0
II. $2{y}^{2}$ - 3y + l = 0
 A). x>y$x>y$ B). x≥y$x\ge y$ C). x
 Answer : Option D Explanation : I. 2x2$2{x}^{2}$ + 2x - x - 1 = 0or, 2x(x + 1) - 1(x + 1) = 0or, (2x - 1) (x + 1) = 0x = 12$\frac{1}{2}$ , -1II. 2y2$2{y}^{2}$ - 2y - y + 1 = 0or, 2y(y - 1) - 1(y - 1) = 0or, (2y - 1)(y - 1) = 0y = 12$\frac{1}{2}$, 1i.e., x≤y$x\le y$

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 : 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 = 8Putting the value of x and y in eqn (i), we getz = 8:. x < y = z

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

8 .I. x = $\sqrt{\left[\left(36{\right)}^{\frac{1}{2}}×\left[1296{\right]}^{\frac{1}{4}}\right]}$
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 : Option D Explanation : x = (62)12×(64)14−−−−−−−−−−−√$\sqrt{\left({6}^{2}{\right)}^{\frac{1}{2}}×\left({6}^{4}{\right)}^{\frac{1}{4}}}$6×6−−−−√$\sqrt{6×6}$ = 6 ..(i)2y + 3z = 33 ... (ii)6y + 5z = 71 ... (iii)From eqn (ii) and (iii),y = 6 and z = 7x = y < z

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 : Option D Explanation : 8x + 7y = 135 ... (i)5x + 6y = 99 ... (ii)9y + 8z = 121 ... (iii)From eqn (i) and (ii),x = 9, and y = 9Putting the value of y in eqn (iii),z = 5:. x = y > z

10 .I. $\left(x+y{\right)}^{3}$ = 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 : Option D Explanation : (x+y)3$\left(x+y{\right)}^{3}$ = 1331or, x + y = 11 ... (i)(x+y)2$\left(x+y{\right)}^{2}$ = 121(x−y)2$\left(x-y{\right)}^{2}$ + 4xy = 121x - y = 3... (ii)[value of xy from eqn (iii)]From eqn (i) and (ii), x = 7, y = 4Put the value x and y in the eqnx - y + z = 07 - y + z = 03 + z = 0z = -3