SBI PO Quadratic Equation Questions and Answers Quiz 2

1 . Directions (Q. Nos. 1-5) : In the following questions two equations numbered 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 the relationship cannot be established $Q.$ I. $\sqrt{1225x} + \sqrt{4900} = 0$ II. $(81)^{1\over 4} y + (343)^{1\over 3} = 0$ A.   $x > y$ B.   $x < y$ C.   $x \geq y$ D.   $x \leq y$ 2 . I. $18\over x^2$ + $6\over x^2$ - $12\over x^2$ = $8\over x^2$ II. $y^ 3$ + 9.68 + 5.64 = 16.95 A.   $x > y$ B.   $x \geq y$ C.   $x \leq y$ D.   x = y or no relation can be established between ‘x’ and ‘y’. 3 . I. $(2)^5 + (11)^3\over 6$ = $x^3$ II. $4y^3 = - (589 \div 4 ) + 5 y^3$ A.   $x > y$ B.   $x \geq y$ C.   $x < y$ D.   $x \leq y$ 4 . I. $12x^ 2$ + llx + 12 = 10x 2 +22x II. $13y^ 2$ - 18y + 3 = 9y 2 - 10y A.   $x > y$ B.   $x \geq y$ C.   $x < y$ D.   $x \leq y$ 5 . I. $(x^{7\over 5} \div 9)$ = $169 \div y{3\over 5}$ II. $y^{1\over 4} \times y^{1\over 4} \times 7$ = $273 \div y^{1\over 2}$ A.   $x > y$ B.   $x \geq y$ C.   $x < y$ D.   $x \leq y$ 6 . Directions (Q. 6 - 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. x = $\sqrt[ 4] {2401}$ II.$2y^ 2$ - 9y - 56 = 0 A.   $x > y$ B.   $x < y$ C.   $x \geq y$ D.   x = y or no relation can be established between ‘x’ and ‘y’. 7 . I. $5x^ 2$ + 3x - 14 = 0 II.$2y^ 2$ - 9y + 10 = 0 A.   $x > y$ B.   $x < y$ C.   $x \geq y$ D.   $x \leq y$ 8 . I. $8x ^ 2$ + 31x + 21 = 0 II. $5y ^2$ + 11y - 36 = 0 A.   $x > y$ B.   $x < y$ C.   $x \leq y$ D.   x = y or no relation can be established between ‘x’ and ‘y’. 9 . I. 3x - y = 12 II. y = $\sqrt{ 1089}$ A.   $x > y$ B.   $x < y$ C.   $x \geq y$ D.   $x \leq y$ 10 . I. $15x^ 2$ + 68x + 77 = 0 II. $3y^ 2$ + 29y + 68 = 0 A.   $x > y$ B.   $x < y$ C.   $x \geq y$ D.   $x \leq y$

Answers & Solutions   1 .     Answer : Option A Explanation : I. $\sqrt{1225x} + \sqrt{4900} = 0$ 35x + 70 = 0 x = - $70\over 35$ = -2 II. 3y + 7 = 0 y = - $7\over 3$ $x > y$ 2 .     Answer : Option D Explanation : I. $18 + 6 x - 12\over x^2$ = $8\over x^2$ or , x = $1\over 3$ = 0.333 II. $y^ 2$ = 16.95 - 9.68 - 5.64 = 1.63 y = ±1.277 3 .     Answer : Option A Explanation : I. $x^3 =$ $32 + 1331\over 6$ = $1363\over6$ II. $5y^3 - 4y^3 =$ $589\over 4$ or $y^3 =$ $589\over 4$ $x > y$ 4 .     Answer : Option B Explanation : I. $2x ^2$ - llx + 12 = 0 or, x = 4, $3\over 2$ II. $4y^ 2$ - 8y + 3 = 0 y = $3\over 2$ , $1\over 2$ $x \geq y$ 5 .     Answer : Option D Explanation : $(x^{7\over 5} \div 9)$ = $169 \div x{3\over 5}$ $(x^{7\over 5} \times x{3\over 5}$ $= 169 x 9$(x^{7+3\over 5} = 1521 $x^2$ = 1521 x = ± 39 II. $y ^{1\over 4}$ + $y^{1\over 4}$ + $y^{1\over 2}$ = $273\over 7$ or, y $1\over 4$ + $1\over 4$ + $1\over 2$ = 39 y = 39 x $\leq$ y 6 .     Answer : Option D Explanation : I. x = $\sqrt[ 4] {2401}$ $\Rightarrow$ x = 7 II.$2y ^ 2$ - 16y + 7y - 56 = 0 2y(y - 8) + 7(y - 8) = 0 (2y + 7) (y - 8) = 0 y = 8 , - $7\over 2$ Hence, no relation exists between x and y. 7 .     Answer : Option B Explanation : I. $5x^ 2$ + 10x - 7x - 14 = 0 or, 5x(x + 2) - 7(x + 2) = 0 or, (x + 2) (5x - 7) = 0 x = - 2, $7\over 5$ II. $2y ^ 2$ - 4y - 5y + 10 = 0 or, 2y(y - 2) - 5(y - 2) = 0 or, (2y - 5)(y - 2) = 0 or, y = 2, $5\over 2$ $x < y$ 8 .     Answer : Option D Explanation : I. $8x^ 2$ + 24x + 7x + 21 = 0 or, 8x(x + 3) + 7(x + 3) = 0 or, (x + 3) (8x + 7) = 0 x = - 3, - $7\over 8$ II. $5y^ 2$ + 20y - 9y - 36 = 0 or, 5y(y + 4) - 9(y + 4) = 0 or, (y + 4) (5y - 9) = 0 y = -4, $9\over 5$ Hence, no relation exists between x and y. 9 .     Answer : Option B Explanation : I. y = $\sqrt{ 1089}$ $\Rightarrow$ y = 33 II. x = $12 + y\over 3$ = $12 + 33 \over 3$ = $45\over 3$ = 15 $x < y$ 10 .     Answer : Option A Explanation : I. $15x^ 2$ + 68x + 77 = 0 or, $15x^ 2$ + 35x + 33x + 77 = 0 or, 5x(3x + 7) + 11(3x + 7) = 0 or, (5x + 11) (3x + 7) = 0 x = -$7\over 3$ , -$11\over 5$ II. $3y^ 2$ + 29y + 68 = 0 or, $3y^ 2$ + 12y + 17y + 68 = 0 or, 3y(y + 4) + 17(y + 4) = 0 or, (y + 4) (3y + 17) = 0 y = -4, -$17\over 3$ $x > y$