# IBPSExamGuru | ibps bankers Guru Adda Clerk PO

## Saturday, 18 March 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 $\geq$y
(4) if x $\leq$y
(5) if x = y, or no relation can be established between x and y.

$Q.$
I. $15x^ 2$ - 19x + 6 = 0
II. $6y^ 2$ - 5y + 1 = 0

 A.   $x > y$ B.   $x < y$ C.   $x \geq y$ D.   $x \leq y$
2 . I. x = $\sqrt{172}$
II. $y^ 2$ - 29y + 210 = 0

 A.   $x > y$ B.   $x < y$ C.   $x \geq y$ D.   $x \leq y$
3 . I. $3x^ 2$ - 20x + 32 = 0
II. $2y^ 2$ - 19y + 44 = 0

 A.   $x > y$ B.   $x < y$ C.   $x \geq y$ D.   $x \leq y$
4 . I. 3x + 8y = -2
II. 4x + 18y = 1

 A.   $x > y$ B.   $x < y$ C.   $x \geq y$ D.   $x \leq y$
5 . I. $2x^ 2$ - 15x + 28 = 0
II.$10y^ 2$ - y - 119 = 0

 A.   $x > y$ B.   $x < y$ C.   $x \geq y$ D.   $x \leq y$
6 . Directions (Q. 6-10): 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 $\geq$ y
(3) if x < y
(4) if x $\leq$ y
(5) if x = y or relationship between x and y cannot be established.

$Q.$
I. 676$x^2$ - 1 = 0
II. y = $1\over \sqrt[3]{13824}$

 A.   $x > y$ B.   $x \geq y$ C.   $x < y$ D.   $x \leq y$
7 . I. 8x + 13y = 62
II 13x - 17y + 128 = 0

 A.   $x > y$ B.   $x \geq y$ C.   $x < y$ D.   $x \leq y$
8 . I. $x^ 2$ = 7x
II. $(y + 7)^ 2$ = 0

 A.   $x > y$ B.   $x \geq y$ C.   $x < y$ D.   $x \leq y$
9 . I.$7x^ 2$ + 2x = 120
II. $y^ 2$ + 11y + 30 = 0

 A.   $x > y$ B.   $x \geq y$ C.   $x < y$ D.   $x \leq y$
10 . I.$2x^ 2$ + 5x - 33 = 0
II. $y^ 2$ - y - 6 = 0

 A.   $x \geq y$ B.   $x < y$ C.   $x \leq y$ D.   x = y or no relation can be established between ‘x’ and ‘y’.

1 .
 Answer : Option A Explanation : I. $15x^ 2$ - 10x - 9x + 6 = 0or, 5x(3x - 2) -3(3x - 2) = 0or, (5x - 3) (3x - 2) = 0x = $3\over 5$, $2\over 3$II. $6y ^ 2$ - 3y - 2y + 1 = 0or, 3y(2y - 1) -1(2y - 1) = 0or, (3y - 1)(2y - 1) = 0y = $1\over 3$, $1\over 2$ x > y
2 .
 Answer : Option B Explanation : I. x = $\sqrt{172}$x = 13.11II. $y^ 2$ - 14y - 15y + 210 = 0or, y(y - 14) - 15(y - 14) = 0or, (y - 14) (y - 15) = 0y = 14, 15x < y
3 .
 Answer : Option D Explanation : I. $3x^ 2$ -12x - 8x + 32 = 0or, 3x(x - 4) - 8(x - 4) = 0or, (x - 4) (3x - 8) = 0x = 4 , $8\over 3$II. $2y^ 2$ - 8y - 11y + 44 = 0or, 2y(y - 4) -11(y - 4) = 0or, (y - 4) (2y - 11) = 0y = 4, $11\over 2$$x \leq y$
4 .
 Answer : Option B Explanation : 4 × eqn (I) - 3 × eqn (II),12x + 32y = -812x + 54y = 3-$\,\,\,\,\,\,\,\,$ -$\,\,\,\,\,\,\,\,\,\,$ - .--------------------22y = -11y = $1\over 2$ amd x = -2X < Y
5 .
 Answer : Option C Explanation : I. $2x^ 2$ - 8x - 7x + 28 = 0or, 2x(x - 4) - 7(x - 4) = 0or, (x - 4) (2x - 7) = 0x = 4, $7\over 2$II. $10y^ 2$ - 35y + 34y - 119 = 0or, 5y(2y - 7) + 17(2y - 7) = 0or, (2y - 7)(5y + 17) = 0y = $7\over 2$ , -$17\over 5$x $\geq$ y
6 .
 Answer : Option C Explanation : I. 676$x^2$ - 1 = 0 $x^2$ = $1\over 676$x = ±$1\over 26$II. y = $1\over \sqrt[3]{13824}$y = $1\over 24$ie,. x < y
7 .
 Answer : Option C Explanation : On solving these two equations, we getx = -2, y = 6ie, x < y
8 .
 Answer : Option A Explanation : I. $x^ 2$ = 7xor, $x 2$ - 7x = 0or, x(x - 7) = 0x = 0, 7II. $(y + 7) 2$ = 0or, (y + 7) = 0y = -7ie, x > y
9 .
 Answer : Option A Explanation : I. $7x^ 2$ - 28x + 30x - 120 = 0or, 7x(x - 4) + 30(x - 4) = 0or, (x - 4) (7x + 30) = 0x = 4, -$30\over 7$II. $y^ 2$ + 6y + 5y + 30 = 0or, y(y + 6) + 5(y + 6) = 0or, (y + 5) (y + 6) = 0y = -5, - 6ie,. x > y
10 .
 Answer : Option D Explanation : I. $2x^ 2$ - 6x + 11x - 33 = 0or, 2x(x - 3) + 11(x - 3) = 0or, (2x + 11) (x - 3) = 0x = 3, -$11\over 2$II.$y^ 2$ - 3y + 2y - 6 = 0or, y(y - 3) + 2(y - 3) = 0or, (y + 2)(y - 3) = 0y = - 2, 3i.e no relation exists between x and y