SBI PO Quadratic Equation Questions and Answers Set – 6

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’.

Answers & Solutions   1 .     Answer : Option A Explanation : I. $15x^ 2$ - 10x - 9x + 6 = 0 or, 5x(3x - 2) -3(3x - 2) = 0 or, (5x - 3) (3x - 2) = 0 x = $3\over 5$, $2\over 3$ II. $6y ^ 2$ - 3y - 2y + 1 = 0 or, 3y(2y - 1) -1(2y - 1) = 0 or, (3y - 1)(2y - 1) = 0 y = $1\over 3$, $1\over 2$ x > y 2 .     Answer : Option B Explanation : I. x = $\sqrt{172}$ x = 13.11 II. $y^ 2$ - 14y - 15y + 210 = 0 or, y(y - 14) - 15(y - 14) = 0 or, (y - 14) (y - 15) = 0 y = 14, 15 x < y 3 .     Answer : Option D Explanation : I. $ 3x^ 2$ -12x - 8x + 32 = 0 or, 3x(x - 4) - 8(x - 4) = 0 or, (x - 4) (3x - 8) = 0 x = 4 , $8\over 3$ II. $2y^ 2 $ - 8y - 11y + 44 = 0 or, 2y(y - 4) -11(y - 4) = 0 or, (y - 4) (2y - 11) = 0 y = 4, $11\over 2$ $x \leq y$ 4 .     Answer : Option B Explanation : 4 × eqn (I) - 3 × eqn (II), 12x + 32y = -8 12x + 54y = 3 -$\,\,\,\,\,\,\,\,$ -$\,\,\,\,\,\,\,\,\,\,$ - . ------------------- -22y = -11 y = $1\over 2 $ amd x = -2 X < Y 5 .     Answer : Option C Explanation : I. $ 2x^ 2$ - 8x - 7x + 28 = 0 or, 2x(x - 4) - 7(x - 4) = 0 or, (x - 4) (2x - 7) = 0 x = 4, $7\over 2$ II. $10y^ 2 $ - 35y + 34y - 119 = 0 or, 5y(2y - 7) + 17(2y - 7) = 0 or, (2y - 7)(5y + 17) = 0 y = $7\over 2$ , -$17\over 5$ 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 get x = -2, y = 6 ie, x < y 8 .     Answer : Option A Explanation : I. $ x^ 2$ = 7x or, $x 2$ - 7x = 0 or, x(x - 7) = 0 x = 0, 7 II. $(y + 7) 2$ = 0 or, (y + 7) = 0 y = -7 ie, x > y 9 .     Answer : Option A Explanation : I. $7x^ 2$ - 28x + 30x - 120 = 0 or, 7x(x - 4) + 30(x - 4) = 0 or, (x - 4) (7x + 30) = 0 x = 4, -$30\over 7$ II. $y^ 2 $ + 6y + 5y + 30 = 0 or, y(y + 6) + 5(y + 6) = 0 or, (y + 5) (y + 6) = 0 y = -5, - 6 ie,. x > y 10 .     Answer : Option D Explanation : I. $2x^ 2 $ - 6x + 11x - 33 = 0 or, 2x(x - 3) + 11(x - 3) = 0 or, (2x + 11) (x - 3) = 0 x = 3, -$11\over 2$ II.$ y^ 2$ - 3y + 2y - 6 = 0 or, y(y - 3) + 2(y - 3) = 0 or, (y + 2)(y - 3) = 0 y = - 2, 3 i.e no relation exists between x and y