SBI PO Quadratic Equation Questions and Answers set - 10

1 . Directions (Q. 1-10): 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 no relation between ‘x’ and ‘y’ can be established. $Q.$ I. $3x^ 2$ – 7x – 20 = 0 II. $y^ 2$ – 8y + 16 = 0 A.   $ x > y$ B.   $ x \geq y $ C.   $ x < y$ D.   $ x \leq y$ 2 . I. $x^ 2$ – 72 = 0 II. $y^ 2$ – 9y + 8 = 0 A.   $ x > y$ B.   $ x \geq y $ C.   $ x < y$ D.   x = y or no relation can be established between ‘x’ and ‘y’. 3 . I. $9x^ 2$ – 114x + 361 = 0 II. $y^ 2$ = 36 A.   $ x > y$ B.   $ x \geq y $ C.   $ x < y$ D.   $ x \leq y$ 4 . I. 13x + 17y = 107 II. x – 11y = – 41 A.   $ x > y$ B.   $ x \geq y $ C.   $ x < y$ D.   $ x \leq y$ 5 . I. $9x^ 2$ + 18x + 9 = 0 II. $y^ 2$ – 3y + 2 = 0 A.   $ x > y$ B.   $ x \geq y $ C.   $ x < y$ D.   $ x \leq y$ 6 . I. 4x + 7y = 42 II. 3x - 11y = – l A.   $ x > y$ B.   $ x \geq y $ C.   $ x < y$ D.   $ x \leq y$ 7 . I. $9x^ 2 $ – 29x + 22 = 0 II.$ y^ 2 $ – 7y + 12 = 0 A.   $ x > y$ B.   $ x \geq y $ C.   $ x < y$ D.   $ x \leq y$ 8 . I. $3x^ 2 $ – 4x – 32 = 0 II. $2y^ 2 $ – 17y + 36 = 0 A.   $ x > y$ B.   $ x \geq y $ C.   $ x < y$ D.   $ x \leq y$ 9 . I. $3x^ 2 $ – 19x – 14 = 0 II. $2y^ 2$ + 5y + 3 = 0 A.   $ x > y$ B.   $ x \geq y $ C.   $ x < y$ D.   $ x \leq y$ 10 . I. $x^ 2 $ + 14x + 49 = 0 II. $y^ 2$ + 9y = 0 A.   $ x > y$ B.   $ x \geq y $ C.   $ x < y$ D.   x = y or no relation can be established between ‘x’ and ‘y’.

Answers & Solutions   1 .     Answer : Option D Explanation : I. $3x^ 2$ - 12x + 5x - 20 = 0 or 3x(x - 4) + 5(x - 4) = 0 or (3x + 5) (x - 4) = 0 x = - $5\over 3$, 4 II. $y^ 2$ - 8y + 16 = 0 or $(y - 4)^ 2 $ = 0 (y - 4) = 0 or y = 4 $x \leq y$ 2 .     Answer : Option D Explanation : I. $x^ 2$ - 72 = 0 or $x^ 2$ = 72 x = + 8.485 II.$y^ 2$ - y - 8y + 8 = 0 or y(y - 1) - 8(y - 1) = 0 or (y - 1) (y - 8) = 0 y = 1, 8 3 .     Answer : Option A Explanation : I. $9x^ 2$ - l14x + 361 =0 or $(3x - 19)^ 2$ = 0 3x - 19 = 0 x = $19\over 3$ = 6.33 II. $y^ 2$ = 36 y = ±6 x > y 4 .     Answer : Option C Explanation : I. 13x + 17y = 107 $\longrightarrow$eqn (II) × 13 13x ± 143y = ± 533 -------------------------- 160y = 640 y = $640\over 160$ = 4 x = 11y - 41 x = 44 - 41 = 3 x < y 5 .     Answer : Option C Explanation : I. $9x^ 2$ + 18x + 9 = 0 or $x^ 2$ + 2x + 1 = 0 or $(x + l)^ 2$ = 0 x + 1 = 0, or x = -1 II.$y 2$ - y - 2y + 2 = 0 or y(y - 1) -2(y - 1) = 0 or (y - 1) (y - 2) = 0 y = 1 , 2 x < y 6 .     Answer : Option A Explanation : eqn (I) ×3 - eqn (II) × 4 12x + 21y = 126 12x - 44y = -4 -$\,\,\,\,\,\,$ +$\,\,\,\,\,\,\,\,\,\,$ + . ---------------------- 65y = 130 y = 2 and x = 7 X > Y 7 .     Answer : Option C Explanation : I. $ 9x^ 2$ - 18x - 1 lx + 22 = 0 or 9x(x - 2)- 11(x - 2) = 0 or (x - 2)(9x - 11) = 0 x = 2, $11\over 9$ II. $y^ 2$ - 3y - 4y + 12 - 0 or y(y - 3) - 4(y - 3) = 0 or (y - 3) (y - 4) = 0 y = 3, 4 x < y 8 .     Answer : Option D Explanation : I. $3x^ 2$ - 4x - 32 = 0 or $3x^ 2$ - 12x + 8x - 32 = 0 or 3x(x - 4) + 8(x - 4) = 0 or (3x + 8) (x - 4) = 0 x = 4, -$8\over 3$ II. $2y^ 2$ - 8y - 9y + 36 = 0 or 2y(y - 4) - 9(y - 4) = 0 or (2y - 9) (y - 4) = 0 or (2y - 9) (y - 4) = 0 y = 4, $9\over 2$ $x \leq y$ 9 .     Answer : Option A Explanation : I. $3x^ 2$ - 21x + 2x - 14 = 0 or 3x(x - 7) + 2(x - 7) = 0 or (3x + 2) (x - 7) = 0 x = 7, -$2\over 3$ II. $2y^ 2$ + 5y + 3 = 0 or $ 2y^ 2$ + 2y + 3y + 3 = 0 or 2y(y + 1) + 3(y + 1) = 0 or (2y + 3) (y + 1) = 0 y = -$3\over 2$, -1 x > y 10 .     Answer : Option D Explanation : I. $x^ 2$ + 14x + 49 = 0 or $(x + 7)^ 2$ = 0 x + 7 = 0 or, x = -7 II. $y^ 2$ + 9y = 0 or y(y + 9) = 0 y = 0, -9 ie,. no relation between x and y.