Questions and Answers Type: | MCQ (Multiple Choice Questions). |

Main Topic: | Quantitative Aptitude. |

Quantitative Aptitude Sub-topic: | Equations and Inequalities Aptitude Questions and Answers. |

Number of Questions: | 10 Questions with Solutions. |

**Directions:** 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 \(y\).

- (I). \(3x^2 + 14x + 15 = 0\), (II). \(2y^2 + 18y + 36 = 0\)
- \(x \gt y\)
- \(x \lt y\)
- \(x \ge y\)
- \(x \le y\)
- \(x = y\) or No relation can be established between \(x\) and \(y\)

Answer: (c) \(x \ge y\)

Solution: $$ 3x^2 + 14x + 15 = 0......(1) $$ $$ 3x^2 + 9x + 5x + 15 = 0 $$ $$ 3x \ (x + 3) + 5 \ (x + 3) = 0 $$ $$ (x + 3) \ (3x + 5) = 0 $$ $$ x = -3, \ and \ x = \frac{-5}{3} $$ $$ 2y^2 + 18y + 36 = 0......(2) $$ $$ 2y^2 + 12y + 6y + 36 = 0 $$ $$ 2y \ (y + 6) + 6 \ (y + 6) = 0 $$ $$ (y + 6) \ (2y + 6) = 0 $$ $$ y = -6, \ and \ y = -3 $$ $$ Hence, \ x \ge y $$

Solution: $$ 3x^2 + 14x + 15 = 0......(1) $$ $$ 3x^2 + 9x + 5x + 15 = 0 $$ $$ 3x \ (x + 3) + 5 \ (x + 3) = 0 $$ $$ (x + 3) \ (3x + 5) = 0 $$ $$ x = -3, \ and \ x = \frac{-5}{3} $$ $$ 2y^2 + 18y + 36 = 0......(2) $$ $$ 2y^2 + 12y + 6y + 36 = 0 $$ $$ 2y \ (y + 6) + 6 \ (y + 6) = 0 $$ $$ (y + 6) \ (2y + 6) = 0 $$ $$ y = -6, \ and \ y = -3 $$ $$ Hence, \ x \ge y $$

- (I). \(8x^2 - 10x - 3 = 0\), (II). \(7y^2 - 12y - 4 = 0\)
- \(x \gt y\)
- \(x \lt y\)
- \(x \ge y\)
- \(x \le y\)
- \(x = y\) or No relation can be established between \(x\) and \(y\)

Answer: (e) No relation can be established between \(x\) and \(y\)

Solution: $$ 8x^2 - 10x - 3 = 0......(1) $$ $$ 8x^2 - 12x + 2x - 3 = 0 $$ $$ 4x \ (2x - 3) + 1 \ (2x - 3) = 0 $$ $$ (2x - 3) \ (4x + 1) = 0 $$ $$ x = \frac{3}{2}, \ and \ x = \frac{-1}{4} $$ $$ 7y^2 - 12y - 4 = 0......(2) $$ $$ 7y^2 - 14y + 2y - 4 = 0 $$ $$ 7y \ (y - 2) + 2 \ (y - 2) = 0 $$ $$ (y - 2) \ (7y + 2) = 0 $$ $$ y = 2, \ and \ y = \frac{-2}{7} $$ Hence, No relation can be established.

Solution: $$ 8x^2 - 10x - 3 = 0......(1) $$ $$ 8x^2 - 12x + 2x - 3 = 0 $$ $$ 4x \ (2x - 3) + 1 \ (2x - 3) = 0 $$ $$ (2x - 3) \ (4x + 1) = 0 $$ $$ x = \frac{3}{2}, \ and \ x = \frac{-1}{4} $$ $$ 7y^2 - 12y - 4 = 0......(2) $$ $$ 7y^2 - 14y + 2y - 4 = 0 $$ $$ 7y \ (y - 2) + 2 \ (y - 2) = 0 $$ $$ (y - 2) \ (7y + 2) = 0 $$ $$ y = 2, \ and \ y = \frac{-2}{7} $$ Hence, No relation can be established.

- (I). \(x^2 + 5x - 6 = 0\), (II). \(y^2 - 6y + 9 = 0\)
- \(x \gt y\)
- \(x \lt y\)
- \(x \ge y\)
- \(x \le y\)
- \(x = y\) or No relation can be established between \(x\) and \(y\)

Answer: (b) \(x \lt y\)

Solution: $$ x^2 + 5x - 6 = 0......(1) $$ $$ x^2 + 6x - x - 6 = 0 $$ $$ x \ (x + 6) - 1 \ (x + 6) = 0 $$ $$ (x + 6) \ (x - 1) = 0 $$ $$ x = -6, \ and \ x = 1 $$ $$ y^2 - 6y + 9 = 0......(2) $$ $$ y^2 - 3y - 3y + 9 = 0 $$ $$ y \ (y - 3) - 3 \ (y - 3) = 0 $$ $$ y = 3, \ and \ y = 3 $$ $$ Hence, \ x \lt y $$

Solution: $$ x^2 + 5x - 6 = 0......(1) $$ $$ x^2 + 6x - x - 6 = 0 $$ $$ x \ (x + 6) - 1 \ (x + 6) = 0 $$ $$ (x + 6) \ (x - 1) = 0 $$ $$ x = -6, \ and \ x = 1 $$ $$ y^2 - 6y + 9 = 0......(2) $$ $$ y^2 - 3y - 3y + 9 = 0 $$ $$ y \ (y - 3) - 3 \ (y - 3) = 0 $$ $$ y = 3, \ and \ y = 3 $$ $$ Hence, \ x \lt y $$

- (I). \(x^2 = 64\), (II). \(y = \sqrt{64}\)
- \(x \gt y\)
- \(x \lt y\)
- \(x \ge y\)
- \(x \le y\)
- \(x = y\) or No relation can be established between \(x\) and \(y\)

Answer: (d) \(x \le y\)

Solution: $$ x^2 = 64......(1) $$ $$ x = \pm 8 $$ $$ y = \sqrt{64}......(2) $$ $$ y = 8 $$ $$ Hence, \ x \le y $$

Solution: $$ x^2 = 64......(1) $$ $$ x = \pm 8 $$ $$ y = \sqrt{64}......(2) $$ $$ y = 8 $$ $$ Hence, \ x \le y $$

- (I). \(15x - 12y = 18\), (II). \(5x + 6y = 8\)
- \(x \gt y\)
- \(x \lt y\)
- \(x \ge y\)
- \(x \le y\)
- \(x = y\) or No relation can be established between \(x\) and \(y\)

Answer: (b) \(x \lt y\)

Solution: $$ 15x - 12y = 18......(1) $$ $$ 5x + 6y = 8......(2) $$ By multiplying equation (2) with 3, we get $$ 15x + 18y = 24......(3) $$ Now, by adding equations (1) and (3), we get $$ 15x - 12y + 15x + 18y = 18 + 24 $$ $$ 6y = 42 $$ $$ y = 7 $$ Now, by putting the value of y in equation, we get $$ 15x - 12y = 18......(1) $$ $$ 15x - 12 \times 7 = 18 $$ $$ 15x - 84 = 18 $$ $$ 15x = 102 $$ $$ x = 6.8 $$ $$ Hence, \ x \lt y $$

Solution: $$ 15x - 12y = 18......(1) $$ $$ 5x + 6y = 8......(2) $$ By multiplying equation (2) with 3, we get $$ 15x + 18y = 24......(3) $$ Now, by adding equations (1) and (3), we get $$ 15x - 12y + 15x + 18y = 18 + 24 $$ $$ 6y = 42 $$ $$ y = 7 $$ Now, by putting the value of y in equation, we get $$ 15x - 12y = 18......(1) $$ $$ 15x - 12 \times 7 = 18 $$ $$ 15x - 84 = 18 $$ $$ 15x = 102 $$ $$ x = 6.8 $$ $$ Hence, \ x \lt y $$

- (I). \(6x^2 - 9x - 6 = 0\), (II). \(4y^2 - 4y - 3 = 0\)
- \(x \gt y\)
- \(x \lt y\)
- \(x \ge y\)
- \(x \le y\)
- \(x = y\) or No relation can be established between \(x\) and \(y\)

Answer: (c) \(x \ge y\)

Solution: $$ 6x^2 - 9x - 6 = 0......(1) $$ $$ 6x^2 - 12x + 3x - 6 = 0 $$ $$ 6x \ (x - 2) + 3 \ (x - 2) = 0 $$ $$ (x - 2) \ (6x + 3) = 0 $$ $$ x = 2, \ and \ x = \frac{-1}{2} $$ $$ 4y^2 - 4y - 3 = 0......(2) $$ $$ 4y^2 - 6y + 2y - 3 = 0 $$ $$ 2y \ (2y - 3) + 1 \ (2y - 3) = 0 $$ $$ (2y - 3) \ (2y + 1) = 0 $$ $$ y = \frac{3}{2}, \ and \ y = \frac{-1}{2} $$ $$ Hence, \ x \ge y $$

Solution: $$ 6x^2 - 9x - 6 = 0......(1) $$ $$ 6x^2 - 12x + 3x - 6 = 0 $$ $$ 6x \ (x - 2) + 3 \ (x - 2) = 0 $$ $$ (x - 2) \ (6x + 3) = 0 $$ $$ x = 2, \ and \ x = \frac{-1}{2} $$ $$ 4y^2 - 4y - 3 = 0......(2) $$ $$ 4y^2 - 6y + 2y - 3 = 0 $$ $$ 2y \ (2y - 3) + 1 \ (2y - 3) = 0 $$ $$ (2y - 3) \ (2y + 1) = 0 $$ $$ y = \frac{3}{2}, \ and \ y = \frac{-1}{2} $$ $$ Hence, \ x \ge y $$

- (I). \(x = \sqrt{1296}\), (II). \(\sqrt{y} = 6\)
- \(x \gt y\)
- \(x \lt y\)
- \(x \ge y\)
- \(x \le y\)
- \(x = y\) or No relation can be established between \(x\) and \(y\)

Answer: (e) \(x = y\)

Solution: $$ x = \sqrt{1296}......(1) $$ $$ x = 36 $$ $$ \sqrt{y} = 6......(2) $$ $$ y = 6^2 $$ $$ y = 36 $$ $$ Hence, \ x = y $$

Solution: $$ x = \sqrt{1296}......(1) $$ $$ x = 36 $$ $$ \sqrt{y} = 6......(2) $$ $$ y = 6^2 $$ $$ y = 36 $$ $$ Hence, \ x = y $$

- (I). \(2x + 2y = 4\), (II). \(x - 2y = 5\)
- \(x \gt y\)
- \(x \lt y\)
- \(x \ge y\)
- \(x \le y\)
- \(x = y\) or No relation can be established between \(x\) and \(y\)

Answer: (a) \(x \gt y\)

Solution: $$ 2x + 2y = 4......(1) $$ $$ x - 2y = 5......(2) $$ By adding equations (1) and (2), we get $$ 2x + 2y + x - 2y = 4 + 5 $$ $$ 3x = 9 $$ $$ x = 3 $$ Now, by putting the value of x in equation (1), we get $$ 2x + 2y = 4......(1) $$ $$ 2 \times 3 + 2y = 4 $$ $$ 6 + 2y = 4 $$ $$ 2y = -2 $$ $$ y = -1 $$ $$ Hence, \ x \gt y $$

Solution: $$ 2x + 2y = 4......(1) $$ $$ x - 2y = 5......(2) $$ By adding equations (1) and (2), we get $$ 2x + 2y + x - 2y = 4 + 5 $$ $$ 3x = 9 $$ $$ x = 3 $$ Now, by putting the value of x in equation (1), we get $$ 2x + 2y = 4......(1) $$ $$ 2 \times 3 + 2y = 4 $$ $$ 6 + 2y = 4 $$ $$ 2y = -2 $$ $$ y = -1 $$ $$ Hence, \ x \gt y $$

- (I). \(9x + 4y = 18\), (II). \(5x + 4y = 6\)
- \(x \gt y\)
- \(x \lt y\)
- \(x \ge y\)
- \(x \le y\)
- \(x = y\) or No relation can be established between \(x\) and \(y\)

Answer: (a) \(x \gt y\)

Solution: $$ 9x + 4y = 18......(1) $$ $$ 5x + 4y = 6......(2) $$ By subtracting equation (2) with equation (1), we get $$ 9x + 4y - (5x + 4y) = 18 - 6 $$ $$ 9x + 4y - 5x - 4y = 12 $$ $$ 4x = 12 $$ $$ x = 3 $$ Now, by putting the value of x in equation (1), we get $$ 9x + 4y = 18......(1) $$ $$ 9 \times 3 + 4y = 18 $$ $$ 27 + 4y = 18 $$ $$ 4y = -9 $$ $$ y = \frac{-9}{4} $$ $$ y = -2.25 $$ $$ Hence, \ x \gt y $$

Solution: $$ 9x + 4y = 18......(1) $$ $$ 5x + 4y = 6......(2) $$ By subtracting equation (2) with equation (1), we get $$ 9x + 4y - (5x + 4y) = 18 - 6 $$ $$ 9x + 4y - 5x - 4y = 12 $$ $$ 4x = 12 $$ $$ x = 3 $$ Now, by putting the value of x in equation (1), we get $$ 9x + 4y = 18......(1) $$ $$ 9 \times 3 + 4y = 18 $$ $$ 27 + 4y = 18 $$ $$ 4y = -9 $$ $$ y = \frac{-9}{4} $$ $$ y = -2.25 $$ $$ Hence, \ x \gt y $$

- (I). \(3x^2 - 5x - 12 = 0\), (II). \(5y^2 + 9y + 4 = 0\)
- \(x \gt y\)
- \(x \lt y\)
- \(x \ge y\)
- \(x \le y\)
- \(x = y\) or No relation can be established between \(x\) and \(y\)

Answer: (e) No relation can be established between \(x\) and \(y\)

Solution: $$ 3x^2 - 5x - 12 = 0......(1) $$ $$ 3x^2 - 9x + 4x - 12 = 0 $$ $$ 3x \ (x - 3) + 4 \ (x - 3) = 0 $$ $$ (x - 3) \ (3x + 4) = 0 $$ $$ x = 3, \ and \ x = \frac{-4}{3} $$ $$ 5y^2 + 9y + 4 = 0......(2) $$ $$ 5y^2 + 5y + 4y + 4 = 0 $$ $$ 5y \ (y + 1) + 4 \ (y + 1) = 0 $$ $$ (y + 1) \ (5y + 4) = 0 $$ $$ y = -1, \ and \ y = \frac{-4}{5} $$ Hence, No relation can be established.

Solution: $$ 3x^2 - 5x - 12 = 0......(1) $$ $$ 3x^2 - 9x + 4x - 12 = 0 $$ $$ 3x \ (x - 3) + 4 \ (x - 3) = 0 $$ $$ (x - 3) \ (3x + 4) = 0 $$ $$ x = 3, \ and \ x = \frac{-4}{3} $$ $$ 5y^2 + 9y + 4 = 0......(2) $$ $$ 5y^2 + 5y + 4y + 4 = 0 $$ $$ 5y \ (y + 1) + 4 \ (y + 1) = 0 $$ $$ (y + 1) \ (5y + 4) = 0 $$ $$ y = -1, \ and \ y = \frac{-4}{5} $$ Hence, No relation can be established.