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IBPS EXAM Guru

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 $\geq$y
(3) if x < y
(4) if x $\leq$y
(5) if x = y or a relationship between x and y cannot be established.

$Q.$
I. $x^ 2$ + 3x = 28
II. $y^ 2$ + 16y + 63 = 0

 A.   $x > y$ B.   $x \geq y$ C.   $x < y$ D.   $x \leq y$
2 . I. x = $\sqrt[3]{2197}$
II. y$^2$ = 169

 A.   $x > y$ B.   $x \geq y$ C.   $x < y$ D.   $x \leq y$
3 . I. $8x^ 2$ - 49x + 45 = 0
II.$8y^ 2$ - y - 9 = 0

 A.   $x > y$ B.   $x \geq y$ C.   $x < y$ D.   $x \leq y$
4 . I. 42x - 17y = -67
II. 7x + 12y = -26

 A.   $x > y$ B.   $x \geq y$ C.   $x < y$ D.   $x \leq y$
5 . I. $x^ 2$ - 8x + 15 = 0
II. $2y^ 2$ - 21y + 55 = 0

 A.   $x > y$ B.   $x \geq y$ C.   $x < y$ 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 p > q
(2) if p $\geq$ q
(3) if p < q
(4) if p $\leq$ q
(5) if p = q or no relation can be established between p and q.

$Q.$
I. $2.3p - 20.01 = 0$
II. $2.9q - p = 0$

 A.   $p > q$ B.   $p \geq q$ C.   $p < q$ D.   $p \leq q$
7 . I. p = $\sqrt{1764}$
II. $q^2$ = 1764

 A.   $p > q$ B.   $p \geq q$ C.   $p < q$ D.   $p \leq q$
8 . I. $p^ 2$ - 26p + 168 = 0
II. $q^ 2$ - 25q + 156 = 0

 A.   p = q or no relation can be established between ‘p’ and ‘q’. B.   $p \leq q$ C.   $p < q$ D.   $p \geq q$
9 . I. $p^ 2$ - 13p + 42 = 0
II. $q^ 2$ + q - 42 = 0

 A.   $p > q$ B.   $p \geq q$ C.   $p < q$ D.   $p \leq q$
10 . I. 6p - 5q = -47
II. 5p + 3q = 11

 A.   $p > q$ B.   $p \geq q$ C.   $p < q$ D.   $p \leq q$

1 .
 Answer : Option B Explanation : I. $x^ 2$ + 3x - 28 = 0or, $x^ 2$ + 7x - 4x - 28 = 0or, x (x + 7) - 4 (x + 7) = 0or, (x - 4) (x + 7) = 0or, x = 4, -7II. $y^ 2$ + 9y + 7y + 63 = 0or, y(y + 9) + 7(y + 9) = 0or, (y + 7)(y + 9) = 0or, y = -7, -9$x \geq y$
2 .
 Answer : Option B Explanation : I. x = $\sqrt[3]{2197}$ x = 13II. y$^2$ = 169y = ± 13x $\geq$ y
3 .
 Answer : Option B Explanation : I. $8x^ 2$ - 40x - 9x + 45 = 0or, 8x (x - 5) -9 (x - 5) = 0or, (8x - 9) (x - 5) = 0or, x = 5, 9/8II. $8y^ 2$ + 8y - 9y -9 = 0or, 8y (y + 1) -9 (y + 1) = 0or, (8y - 9) (y + 1) = 0y = $9\over 8$, -1x$\geq$y
4 .
 Answer : Option C Explanation : 42x - 17y = -6742x + 72y = -156$\,\,\,\,\,$$\longrightarrow$ eqn (II) × 6-$\,\,\,\,\,\,\,$ -$\,\,\,\,\,\,\,\,\,\,\,\,\,$ + -----------------------89y = 89y = -$89\over 89$ = -1 & x = -2x < y
5 .
 Answer : Option D Explanation : I. $x^ 2$ - 8x + 15 = 0or, $x^ 2$ - 3x - 5x + 15 = 0or, x (x - 3) -5 (x - 3) = 0or, (x - 3) (x - 5) = 0or, x = 3, 5II. $2y^ 2$ - 10y + 55 = 0or, 2y (y - 5) -11 (y - 5) = 0or, (y - 5)(2y - 11) = 0y = 5, $11\over 2$x $\leq$ y
6 .
 Answer : Option A Explanation : I. 2.3p - 20.01 = 0p = $20.01\over 2.3$ = 8.7II. $2.9q - p$ = 0 or, p = 2.9qq = $8.7\over 2.9$ = 8.7ie,. p > q
7 .
 Answer : Option B Explanation : I. p = $\sqrt{1764}$ p = 42II. $q^2$ = 1764q = ± 42P$\geq$ q
8 .
 Answer : Option A Explanation : I.$p^ 2$ - 26p + 168 = 0$p^ 2$ - 12p - 14p + 168 = 0p(p - 12) - 14(p - 12) = 0(p - 12) (p - 14) = 0p = 12, 14II. $q^ 2$ - 25q + 156 = 0$q^ 2$ - 13q - 12q + 156 = 0q(q - 13) - 12(q - 13) = 0(q - 12) (q - 13) = 0q = 12, 13Hence, no relation can be established between p and q
9 .
 Answer : Option B Explanation : I. $p^ 2$ - l3q + 42 = 0$p^ 2$ - 6p - 7p + 42 = 0p(p - 6) - 7(p - 6) = 0(p - 6) (p - 7) = 0p = 6, 7II. $q^ 2$ + q - 42 = 0$q^ 2$ + 7q - 6p - 42 = 0q(q + 7) - 6(q + 7) = 0(q - 6)(q + 7) = 0q = 6, - 7 ie p $\geq$ q
10 .
 Answer : Option C Explanation : eqn(I) × 3 $\,\,\,\,\,\,$ 18p - 15q = -141eqn (II) × 5 $\,\,\,\,$ 25p + 15q = 55$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$-----------------------$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ 43p = - 86 p = -$86\over 43$ = -25p + 3q = 113q = 11 - 5p3q = 11 + 103q = 21q = 7 ie p < q