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). \(x^2 = 36\), (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: (e) No relation can be established between \(x\) and \(y\)

Solution: $$ x^2 = 36......(1) $$ $$ x = \pm 36 $$ $$ y^2 - 6y + 9 = 0......(2) $$ $$ y^2 - 3y - 3y + 9 = 0 $$ $$ y \ (y - 3) - 3 \ (y - 3) = 0 $$ $$ (y - 3) \ (y - 3) = 0 $$ $$ y = 3 $$ Hence, No relation can be established.

Solution: $$ x^2 = 36......(1) $$ $$ x = \pm 36 $$ $$ y^2 - 6y + 9 = 0......(2) $$ $$ y^2 - 3y - 3y + 9 = 0 $$ $$ y \ (y - 3) - 3 \ (y - 3) = 0 $$ $$ (y - 3) \ (y - 3) = 0 $$ $$ y = 3 $$ Hence, No relation can be established.

- (I). \(2x^2 - 8x - 24 = 0\), (II). \(5y^2 + 10y + 5 = 0\)
- \(x \gt y\)
- \(x \lt y\)
- \(x \ge y\)
- \(x \le y\)
- \(x = y\) or No relation can be established between \(x\) and \(y\)

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

Solution: $$ 2x^2 - 8x - 24 = 0......(1) $$ $$ 2x^2 - 12x + 4x - 24 = 0 $$ $$ 2x \ (x - 6) + 4 \ (x - 6) = 0 $$ $$ (x - 6) \ (2x + 4) = 0 $$ $$ x = 6, \ and \ x = -2 $$ $$ 5y^2 + 10y + 5 = 0......(2) $$ $$ 5y^2 + 5y + 5y + 5 = 0 $$ $$ 5y \ (y + 1) + 5 \ (y + 1) = 0 $$ $$ (y + 1) \ (5y + 5) = 0 $$ $$ y = -1 $$ Hence, No relation can be established.

Solution: $$ 2x^2 - 8x - 24 = 0......(1) $$ $$ 2x^2 - 12x + 4x - 24 = 0 $$ $$ 2x \ (x - 6) + 4 \ (x - 6) = 0 $$ $$ (x - 6) \ (2x + 4) = 0 $$ $$ x = 6, \ and \ x = -2 $$ $$ 5y^2 + 10y + 5 = 0......(2) $$ $$ 5y^2 + 5y + 5y + 5 = 0 $$ $$ 5y \ (y + 1) + 5 \ (y + 1) = 0 $$ $$ (y + 1) \ (5y + 5) = 0 $$ $$ y = -1 $$ Hence, No relation can be established.

- (I). \(15x^2 + 26x + 8 = 0\), (II). \(4y^2 - 14y - 8 = 0\)
- \(x \gt y\)
- \(x \lt y\)
- \(x \ge y\)
- \(x \le y\)
- \(x = y\) or No relation can be established between \(x\) and \(y\)

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

Solution: $$ 15x^2 + 26x + 8 = 0......(1) $$ $$ 15x^2 + 20x + 6x + 8 = 0 $$ $$ 5x \ (3x + 4) + 2 \ (3x + 4) = 0 $$ $$ (5x + 2) \ (3x + 4) = 0 $$ $$ x = \frac{-2}{5}, \ and \ x = \frac{-4}{3} $$ $$ 4y^2 - 14y - 8 = 0......(2) $$ $$ 4y^2 - 16y + 2y - 8 = 0 $$ $$ 4y \ (y - 4) + 2 \ (y - 4) = 0 $$ $$ (y - 4) \ (4y + 2) = 0 $$ $$ y = 4, \ and \ y = \frac{-1}{2} $$ Hence, No relation can be established.

Solution: $$ 15x^2 + 26x + 8 = 0......(1) $$ $$ 15x^2 + 20x + 6x + 8 = 0 $$ $$ 5x \ (3x + 4) + 2 \ (3x + 4) = 0 $$ $$ (5x + 2) \ (3x + 4) = 0 $$ $$ x = \frac{-2}{5}, \ and \ x = \frac{-4}{3} $$ $$ 4y^2 - 14y - 8 = 0......(2) $$ $$ 4y^2 - 16y + 2y - 8 = 0 $$ $$ 4y \ (y - 4) + 2 \ (y - 4) = 0 $$ $$ (y - 4) \ (4y + 2) = 0 $$ $$ y = 4, \ and \ y = \frac{-1}{2} $$ Hence, No relation can be established.

- (I). \(20x^2 - 16x + 3 = 0\), (II). \(12y^2 - 12y + 3 = 0\)
- \(x \gt y\)
- \(x \lt y\)
- \(x \ge y\)
- \(x \le y\)
- \(x = y\) or No relation can be established between \(x\) and \(y\)

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

Solution: $$ 20x^2 - 16x + 3 = 0......(1) $$ $$ 20x^2 - 10x - 6x + 3 = 0 $$ $$ 10x \ (2x - 1) - 3 \ (2x - 1) = 0 $$ $$ (2x - 1) \ (10x - 3) = 0 $$ $$ x = \frac{1}{2}, \ and \ x = \frac{3}{10} $$ $$ 12y^2 - 12y + 3 = 0......(2) $$ $$ 12y^2 - 6y - 6y + 3 = 0 $$ $$ 6y \ (2y - 1) - 3 \ (2y - 1) = 0 $$ $$ (2y - 1) \ (6y - 3) = 0 $$ $$ y = \frac{1}{2} $$ $$ Hence, \ x \le y $$

Solution: $$ 20x^2 - 16x + 3 = 0......(1) $$ $$ 20x^2 - 10x - 6x + 3 = 0 $$ $$ 10x \ (2x - 1) - 3 \ (2x - 1) = 0 $$ $$ (2x - 1) \ (10x - 3) = 0 $$ $$ x = \frac{1}{2}, \ and \ x = \frac{3}{10} $$ $$ 12y^2 - 12y + 3 = 0......(2) $$ $$ 12y^2 - 6y - 6y + 3 = 0 $$ $$ 6y \ (2y - 1) - 3 \ (2y - 1) = 0 $$ $$ (2y - 1) \ (6y - 3) = 0 $$ $$ y = \frac{1}{2} $$ $$ Hence, \ x \le y $$

- (I). \(24x^2 - 14x - 5 = 0\), (II). \(18y^2 - 17y + 4 = 0\)
- \(x \gt y\)
- \(x \lt y\)
- \(x \ge y\)
- \(x \le y\)
- \(x = y\) or No relation can be established between \(x\) and \(y\)

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

Solution: $$ 24x^2 - 14x - 5 = 0......(1) $$ $$ 24x^2 - 20x + 6x - 5 = 0 $$ $$ 4x \ (6x - 5) + 1 \ (6x - 5) = 0 $$ $$ (6x - 5) \ (4x + 1) = 0 $$ $$ x = \frac{5}{6}, \ and \ x = \frac{-1}{4} $$ $$ 18y^2 - 17y + 4 = 0......(2) $$ $$ 18y^2 - 9y - 8y + 4 = 0 $$ $$ 9y \ (2y - 1) - 4 \ (2y - 1) = 0 $$ $$ (2y - 1) \ (9y - 4) = 0 $$ $$ y = \frac{1}{2}, \ and \ y = \frac{4}{9} $$ Hence, No relation can be established.

Solution: $$ 24x^2 - 14x - 5 = 0......(1) $$ $$ 24x^2 - 20x + 6x - 5 = 0 $$ $$ 4x \ (6x - 5) + 1 \ (6x - 5) = 0 $$ $$ (6x - 5) \ (4x + 1) = 0 $$ $$ x = \frac{5}{6}, \ and \ x = \frac{-1}{4} $$ $$ 18y^2 - 17y + 4 = 0......(2) $$ $$ 18y^2 - 9y - 8y + 4 = 0 $$ $$ 9y \ (2y - 1) - 4 \ (2y - 1) = 0 $$ $$ (2y - 1) \ (9y - 4) = 0 $$ $$ y = \frac{1}{2}, \ and \ y = \frac{4}{9} $$ Hence, No relation can be established.

- (I). \(32x^2 + 24x + 4 = 0\), (II). \(y = \sqrt{25}\)
- \(x \gt y\)
- \(x \lt y\)
- \(x \ge y\)
- \(x \le y\)
- \(x = y\) or No relation can be established between \(x\) and \(y\)

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

Solution: $$ 32x^2 + 24x + 4 = 0......(1) $$ $$ 32x^2 + 16x + 8x + 4 = 0 $$ $$ 16x \ (2x + 1) + 4 \ (2x + 1) = 0 $$ $$ (2x + 1) \ (16x + 4) = 0 $$ $$ x = \frac{-1}{2}, \ and \ x = \frac{-1}{4} $$ $$ y = \sqrt{25}......(2) $$ $$ y = 5 $$ $$ Hence, \ x \lt y $$

Solution: $$ 32x^2 + 24x + 4 = 0......(1) $$ $$ 32x^2 + 16x + 8x + 4 = 0 $$ $$ 16x \ (2x + 1) + 4 \ (2x + 1) = 0 $$ $$ (2x + 1) \ (16x + 4) = 0 $$ $$ x = \frac{-1}{2}, \ and \ x = \frac{-1}{4} $$ $$ y = \sqrt{25}......(2) $$ $$ y = 5 $$ $$ Hence, \ x \lt y $$

- (I). \(7x^2 - 18x + 8 = 0\), (II). \(9y^2 + 18y + 5 = 0\)
- \(x \gt y\)
- \(x \lt y\)
- \(x \ge y\)
- \(x \le y\)
- \(x = y\) or No relation can be established between \(x\) and \(y\)

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

Solution: $$ 7x^2 - 18x + 8 = 0......(1) $$ $$ 7x^2 - 14x - 4x + 8 = 0 $$ $$ 7x \ (x - 2) - 4 \ (x - 2) = 0 $$ $$ (x - 2) \ (7x - 4) = 0 $$ $$ x = 2, \ and \ x = \frac{4}{7} $$ $$ 9y^2 + 18y + 5 = 0......(2) $$ $$ 9y^2 + 15y + 3y + 5 = 0 $$ $$ 3y \ (3y + 5) + 1 \ (3y + 5) = 0 $$ $$ (3y + 5) \ (3y + 1) = 0 $$ $$ y = \frac{-5}{3}, \ and \ y = \frac{-1}{3} $$ $$ Hence, \ x \gt y $$

Solution: $$ 7x^2 - 18x + 8 = 0......(1) $$ $$ 7x^2 - 14x - 4x + 8 = 0 $$ $$ 7x \ (x - 2) - 4 \ (x - 2) = 0 $$ $$ (x - 2) \ (7x - 4) = 0 $$ $$ x = 2, \ and \ x = \frac{4}{7} $$ $$ 9y^2 + 18y + 5 = 0......(2) $$ $$ 9y^2 + 15y + 3y + 5 = 0 $$ $$ 3y \ (3y + 5) + 1 \ (3y + 5) = 0 $$ $$ (3y + 5) \ (3y + 1) = 0 $$ $$ y = \frac{-5}{3}, \ and \ y = \frac{-1}{3} $$ $$ Hence, \ x \gt y $$

- (I). \(2x^2 - 72 = 0\), (II). \(y = \sqrt{36}\)
- \(x \gt y\)
- \(x \lt y\)
- \(x \ge y\)
- \(x \le y\)
- \(x = y\) or No relation can be established between \(x\) and \(y\)

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

Solution: $$ 2x^2 - 72 = 0......(1) $$ $$ 2x^2 = 72 $$ $$ x^2 = 36 $$ $$ x = \pm 6 $$ $$ y = \sqrt{36}......(2) $$ $$ y = 6 $$ $$ Hence, \ x \le y $$

Solution: $$ 2x^2 - 72 = 0......(1) $$ $$ 2x^2 = 72 $$ $$ x^2 = 36 $$ $$ x = \pm 6 $$ $$ y = \sqrt{36}......(2) $$ $$ y = 6 $$ $$ Hence, \ x \le y $$

- (I). \(x^2 - \sqrt{25} = \sqrt{16}\), (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: (d) \(x \le y\)

Solution: $$ x^2 - \sqrt{25} = \sqrt{16}......(1) $$ $$ x^2 = \sqrt{16} + \sqrt{25} $$ $$ x^2 = 4 + 5 $$ $$ x^2 = 9 $$ $$ x = \pm 3 $$ $$ y^2 - 6y + 9 = 0......(2) $$ $$ y^2 - 3y - 3y + 9 = 0 $$ $$ y \ (y - 3) - 3 \ (y - 3) = 0 $$ $$ (y - 3) \ (y - 3) = 0 $$ $$ y = 3 $$ $$ Hence, \ x \le y $$

Solution: $$ x^2 - \sqrt{25} = \sqrt{16}......(1) $$ $$ x^2 = \sqrt{16} + \sqrt{25} $$ $$ x^2 = 4 + 5 $$ $$ x^2 = 9 $$ $$ x = \pm 3 $$ $$ y^2 - 6y + 9 = 0......(2) $$ $$ y^2 - 3y - 3y + 9 = 0 $$ $$ y \ (y - 3) - 3 \ (y - 3) = 0 $$ $$ (y - 3) \ (y - 3) = 0 $$ $$ y = 3 $$ $$ Hence, \ x \le y $$

- (I). \(2x^2 - 6x + 4 = 0\), (II). \(y^2 - 12y + 32 = 0\)
- \(x \gt y\)
- \(x \lt y\)
- \(x \ge y\)
- \(x \le y\)
- \(x = y\) or No relation can be established between \(x\) and \(y\)

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

Solution: $$ 2x^2 - 6x + 4 = 0......(1) $$ $$ 2x^2 - 4x - 2x + 4 = 0 $$ $$ 2x \ (x - 2) - 2 \ (x - 2) = 0 $$ $$ (x - 2) \ (2x - 2) = 0 $$ $$ x = 2, \ and \ x = 1 $$ $$ y^2 - 12y + 32 = 0......(2) $$ $$ y^2 - 8y - 4y + 32 = 0 $$ $$ y \ (y - 8) - 4 \ (y - 8) = 0 $$ $$ (y - 8) \ (y - 4) = 0 $$ $$ y = 8, \ and \ y = 4 $$ $$ Hence, \ x \lt y $$

Solution: $$ 2x^2 - 6x + 4 = 0......(1) $$ $$ 2x^2 - 4x - 2x + 4 = 0 $$ $$ 2x \ (x - 2) - 2 \ (x - 2) = 0 $$ $$ (x - 2) \ (2x - 2) = 0 $$ $$ x = 2, \ and \ x = 1 $$ $$ y^2 - 12y + 32 = 0......(2) $$ $$ y^2 - 8y - 4y + 32 = 0 $$ $$ y \ (y - 8) - 4 \ (y - 8) = 0 $$ $$ (y - 8) \ (y - 4) = 0 $$ $$ y = 8, \ and \ y = 4 $$ $$ Hence, \ x \lt y $$