Quadratic Equation Practice set 4 for IBPS Clerk 2017

Dear Aspirants, Here Find Practice on Quadratic Equation Practice set 4 for IBPS Clerk 2017

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 relationship between x and y cannot be established $Q.$ I.11x + 5y = 117 II. 7x + 13y = 153 A). $ x > y$ B). $ x \geq y $ C). $ x < y$ D). $ x \leq y$ Answer & Explanation Answer : Option C Explanation : eqn (I) × 7 eqn (II) × 11 $\,\,\,$77x + 35y = 819 - 77x ± 143y = 1683 ------------------------------ - 108y = - 864 y = 8, x = 7 ie x < y View Answer 2 . I.$6x^ 2$ + 51x + 105 = 0 II. $2y^ 2$ + 25y + 78 = 0 A). $ x > y$ B). $ x \geq y $ C). $ x < y$ D). $ x \leq y$ Answer & Explanation Answer : Option C Explanation : I. $6x^ 2$ + 21x + 30x + 105 = 0 or, 3x(2x + 7) + 15(2x + 7) = 0 or, (3x + 15) (2x + 7) = 0 x = -5 , -$7\over 2$ II. $ 2y^ 2$ + 12y + 13y + 78 = 0 or, 2y(y + 6) + 13(y + 6) = 0 or, (2y + 13) (y + 6) = 0 y = -$13\over 2$ , -6 $x < y$ View Answer 3 . I.6x + 7y = 52 II. 14x + 4y = 35 A). $ x > y$ B). $ x \geq y $ C). $ x < y$ D). $ x \leq y$ Answer & Explanation Answer : Option C Explanation : eqn (I) × 4 eqn (II) × 7 24x + 28y = 208 98x ± 28y = 245 - ---------------------- - 74x = - 37 x = $1\over 2$, y = 7 $x < y$ View Answer 4 . I.$x^ 2$ + 11x + 30 = 0 II. $y^ 2$ + 12y + 36 = 0 A). $ x > y$ B). $ x \geq y $ C). $ x < y$ D). $ x \leq y$ Answer & Explanation Answer : Option B Explanation : I. $x^ 2$ + 5x + 6x + 30 = 0 or, x(x + 5) + 6(x + 5) = 0 or, (x + 5) (x + 6) = 0 x = - 5, - 6 II. $y^ 2$ + 12y + 36 = 0 or, $(y + 6)^ 2$ = 0 or, y + 6 = 0 y = - 6 ie x $\geq$ y View Answer 5 . I.$2x^ 2$ + x - 1 = 0 II. $2y^ 2$ - 3y + l = 0 A). $ x > y$ B). $ x \geq y $ C). $ x < y$ D). $ x \leq y$ Answer & Explanation Answer : Option D Explanation : I. $2x^ 2$ + 2x - x - 1 = 0 or, 2x(x + 1) - 1(x + 1) = 0 or, (2x - 1) (x + 1) = 0 x = $1\over 2$ , -1 II. $2y^ 2$ - 2y - y + 1 = 0 or, 2y(y - 1) - 1(y - 1) = 0 or, (2y - 1)(y - 1) = 0 y = $1\over 2$, 1 i.e., $x \leq y$ View Answer 6 . Directions (Q.6-10) : In the following questions three equations numbered I, II and III are given. You have to solve all the equations either together or separately, or two together and one separately, or by any other method and give answer If (1) x < y = z (2) x < y < z (3) x < y > z (4) x = y > z (5) x = y = z or if none of the above relationship is established $Q.$ I. 7x + 6y + 4z = 122 II. 4x + 5y + 3z = 88 III. 9x + 2y + z = 78 A). x < y = z B). x < y < z C). x < y > z D). x = y > z Answer & Explanation Answer : Option A Explanation : 7x + 6y + 4z = 122 ... (i) 4x + 5y + 3z = 88 ... (ii) 9x + 2y + z = 78 ... (iii) From (i) and (ii) 5x - 2y = 14... (iv) From (ii) and (iii) 23x + y = 146 ... (v) From (iv) and (v), x = 6, y = 8 Putting the value of x and y in eqn (i), we get z = 8 :. x < y = z View Answer 7 . I. 7x + 6y =110 II. 4x + 3y = 59 III. x + z = 15 A). x < y = z B). x < y < z C). x < y > z D). x = y > z Answer & Explanation Answer : Option C Explanation : 7x + 6y = 110 ... (i) 4x + 3y = 59 ... (ii) x + z = 15 ... (iii) From eqn (i) and (ii), x = 8, y = 9 Put the value of x in eqn (iii). Then, z = 7 x < y > z View Answer 8 . I. x = $\sqrt{[(36)^{1\over 2} \times [1296]^{1\over 4}]}$ II. 2y + 3z = 33 III. 6y + 5z = 71 A). x < y = z B). x < y < z C). x < y > z D). x = y < z Answer & Explanation Answer : Option D Explanation : x = $\sqrt{(6^2)^{1\over 2} \times (6^4)^{1\over 4}}$ $\sqrt{6\times6}$ = 6 ..(i) 2y + 3z = 33 ... (ii) 6y + 5z = 71 ... (iii) From eqn (ii) and (iii), y = 6 and z = 7 x = y < z View Answer 9 . I. 8x + 7y= 135 II. 5x + 6y = 99 III. 9y + 8z = 121 A). x < y = z B). x < y < z C). x < y > z D). x = y > z Answer & Explanation Answer : Option D Explanation : 8x + 7y = 135 ... (i) 5x + 6y = 99 ... (ii) 9y + 8z = 121 ... (iii) From eqn (i) and (ii), x = 9, and y = 9 Putting the value of y in eqn (iii), z = 5 :. x = y > z View Answer 10 . I. $(x + y)^ 3$ = 1331 II. x - y + z = 0 III. xy = 28 A). x < y = z B). x < y < z C). x < y > z D). x = y = z or if none of the above relationship is established Answer & Explanation Answer : Option D Explanation : $(x + y)^ 3$ = 1331 or, x + y = 11 ... (i) $(x + y)^ 2$ = 121 $(x - y)^ 2$ + 4xy = 121 x - y = 3... (ii) [value of xy from eqn (iii)] From eqn (i) and (ii), x = 7, y = 4 Put the value x and y in the eqn x - y + z = 0 7 - y + z = 0 3 + z = 0 z = -3 View Answer