Quadratic Equation Practice set 3 for IBPS Clerk 2017

Dear Aspirants, Take practice on Quadratic Equation Practice set 3 for IBPS Clerk 2017

1 . Directions (Q. 1-10): 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. $2x^ 2$ + x - 1 = 0 II. $6y^ 2$ - 13y + 5 = 0 A). $x > y$ B). $x < y$ C). $x \geq y$ D). x = y or no relation can be established between ‘x’ and ‘y’. Answer & Explanation Answer : Option B Explanation : I. $2x^ 2$ + 2x - x - 1 = 0 or, 2x(x + 1) - 1(x + 1) = 0 or, (x + 1) (2x - 1) = 0 x = -1 , $1\over 2$ II. $6y^ 2$ - 3y - 10y + 5 = 0 or, 3y(2y - 1) - 5(2y - 1) = 0 or, (3y - 5)(2y - 1) = 0 y = -3, $11\over 3$ $x < y$ View Answer 2 . I. $21x ^ 2$ - 122x + 165 = 0 II. $3y^2$ - 2y - 33 = 0 A). $x > y$ B). $x \geq y$ C). $x < y$ D). x = y or no relation can be established between ‘x’ and ‘y’. Answer & Explanation Answer : Option D Explanation : I. $21x ^ 2$ - 45x - 77x + 165 = 0 or, 3x(7x - 15) - 11 (7x - 15) = 0 or, (3x - 11) (7x - 15) = 0 x = $11\over 3$, $15\over 7$ II. $3y^ 2$ + 9y - 11y - 33 = 0 or,3y(y + 3) - 11(y + 3) = 0 or,(3y - 11) (y + 3) = 0 y = -3 , $11\over 3$ Hence, no relation can be established between x and y. View Answer 3 . I. $5x 2$ - 29x + 36 = 0 II. $10y 2$ - 3y - 27 = 0 A). $x > y$ B). $x \leq y$ C). $x < y$ D). $x \geq y$ Answer & Explanation Answer : Option D Explanation : I. $5x^ 2$ - 20x - 9x + 36 = 0 or, 5x(x - 4) - 9(x - 4) = 0 or,(x - 4) (5x - 9) = 0 x = 4 , $9\over 5$ II. $10y ^ 2$ + 15y - 18y - 27 = 0 or, 5y(2y + 3) - 9(2y + 3) = 0 or, (2y + 3) (5y - 9) = 0 y = $9\over 5$, - $3\over 2$ x $\geq$ y View Answer 4 . I. 7x + 4y = 3 II. 5x + 3y = 3 A). $x > y$ B). $x < y$ C). $x \leq y$ D). $x \geq y$ Answer & Explanation Answer : Option B Explanation : eqn (I) × 3 - eqn (II) × 4 21x + 12y = 9 20x + 12y = 12 -$\,\,\,\,\,\,\,\,\,$ -$\,\,\,\,\,\,\,\,\,$ - ------------------- x = - 3 and y = 6 $x < y$ View Answer 5 . I. $7x^ 2$ - 54x + 99 = 0 II. $4y^ 2$ - 16y + 15 = 0 A). $x > y$ B). $x < y$ C). $x \geq y$ D). $x \leq y$ Answer & Explanation Answer : Option A Explanation : I. $7x^ 2$ - 21x - 33x + 99 = 0 or, 7x(x - 3) - 33(x - 3) = 0 or, (x - 3) (7x - 33) = 0 x = 3, $33\over 7$ II. $4y^ 2$ - 6y - 10y + 15 = 0 or, 2y(2y - 3) - 5(2y - 3) = 0 or, (2y - 3)(2y - 5) = 0 y = $3\over 2$, $5\over 2$ $x > y$ View Answer 6 . I. $5x^ 2$ - 87x + 378 = 0 II. $3y^ 2$ - 49y + 200 = 0 A). $x > y$ B). $x < y$ C). $x \geq y$ D). $x \leq y$ Answer & Explanation Answer : Option A Explanation : I. $5x^ 2$ - 45x - 42x + 378 = 0 or, 5x(x - 9) - 42(x - 9) = 0 or, (5x - 42) (x - 9) = 0 x = 9 , $42\over 5$ II. $3y^ 2$ - 24y - 25y + 200 = 0 or, 3y(y - 8) - 25(y - 8) = 0 or, (y - 8) (3y - 25) = 0 y = 8, $25\over 3$ $x > y$ View Answer 7 . I. $10x^ 2$ - x - 24 = 0 II. $y^ 2$ - 2y = 0 A). x = y or no relation can be established between ‘x’ and ‘y’. B). $x < y$ C). $x \geq y$ D). $x \leq y$ Answer & Explanation Answer : Option A Explanation : I. $10x^ 2$ - 16x + 15x - 24 = 0 or, 2x(5x - 8) + 3(5x - 8) = 0 or, (2x + 3) (5x - 8) = 0 x = -$3\over 2$, $8\over5$ II.$y^ 2$ - 2y = 0 or, y(y - 2) = 0 y = 0, 2 ie no relationship exists between x and y View Answer 8 . I. $x^ 2$ - 5x + 6 = 0 II. $2y^ 2$ - 15y + 27 = 0 A). $x > y$ B). $x < y$ C). $x \geq y$ D). $x \leq y$ Answer & Explanation Answer : Option D Explanation : I.$x^ 2$ - 2x - 3x + 6 = 0 or, x(x - 2) - 3(x - 2) = 0 or, (x - 2) (x - 3) = 0 x = 2, 3 II. $2y ^ 2$ - 6y - 9y + 27 = 0 or, 2y(y - 3) - 9(y - 3) = 0 or, (y - 3) (2y - 9) = 0 y = 3, $9\over 2$ $x \leq y$ View Answer 9 . I. 3x + 2y = 301 II. 7x - 5y = 74 A). $x > y$ B). $x < y$ C). $x \geq y$ D). $x \leq y$ Answer & Explanation Answer : Option B Explanation : eqn (I) × 5 + eqn (II) × 2 15x + 10y = 1505 14x - 10y = 148 ------------------------ 29x $\,\,\,\,\,\,\,\,\,\,$ = 1653 $x$ = $1653\over 29$ = 57 and y = 65 $x < y$ View Answer 10 . I. $14x ^ 2$ - 37x + 24 = 0 II. $28y^ 2$ - 53y + 24 = 0 A). $x > y$ B). $x < y$ C). $x \geq y$ D). $x \leq y$ Answer & Explanation Answer : Option C Explanation : I. $14x^ 2$ - 37x + 24 = 0 or, $14x^ 2$ - 21x - 16x + 24 = 0 or, 7x(2x - 3) - 8(2x - 3) = 0 or, (2x - 3) (7x - 8) = 0 x = $3\over 2$, $8\over 7$ II. $28y^ 2$ - 53y + 24 = 0 or, $28y^ 2$ - 21y - 32y + 24 = 0 or, 7y(4y - 3) - 8(4y - 3) = 0 or, (7y - 8) (4y - 3) = 0 y = $8 \over 7$, $3\over 4$ $x \geq y$ View Answer