# Equations and Inequalities Questions and Answers:

#### Overview:

 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 $$y$$.

1. (I). $$x^2 = 36$$, (II). $$y^2 - 6y + 9 = 0$$

1. $$x \gt y$$
2. $$x \lt y$$
3. $$x \ge y$$
4. $$x \le y$$
5. $$x = y$$ 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.

1. (I). $$2x^2 - 8x - 24 = 0$$, (II). $$5y^2 + 10y + 5 = 0$$

1. $$x \gt y$$
2. $$x \lt y$$
3. $$x \ge y$$
4. $$x \le y$$
5. $$x = y$$ 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.

1. (I). $$15x^2 + 26x + 8 = 0$$, (II). $$4y^2 - 14y - 8 = 0$$

1. $$x \gt y$$
2. $$x \lt y$$
3. $$x \ge y$$
4. $$x \le y$$
5. $$x = y$$ 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.

1. (I). $$20x^2 - 16x + 3 = 0$$, (II). $$12y^2 - 12y + 3 = 0$$

1. $$x \gt y$$
2. $$x \lt y$$
3. $$x \ge y$$
4. $$x \le y$$
5. $$x = y$$ or No relation can be established between $$x$$ and $$y$$

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

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

1. (I). $$24x^2 - 14x - 5 = 0$$, (II). $$18y^2 - 17y + 4 = 0$$

1. $$x \gt y$$
2. $$x \lt y$$
3. $$x \ge y$$
4. $$x \le y$$
5. $$x = y$$ 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.

1. (I). $$32x^2 + 24x + 4 = 0$$, (II). $$y = \sqrt{25}$$

1. $$x \gt y$$
2. $$x \lt y$$
3. $$x \ge y$$
4. $$x \le y$$
5. $$x = y$$ or No relation can be established between $$x$$ and $$y$$

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

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

1. (I). $$7x^2 - 18x + 8 = 0$$, (II). $$9y^2 + 18y + 5 = 0$$

1. $$x \gt y$$
2. $$x \lt y$$
3. $$x \ge y$$
4. $$x \le y$$
5. $$x = y$$ or No relation can be established between $$x$$ and $$y$$

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

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

1. (I). $$2x^2 - 72 = 0$$, (II). $$y = \sqrt{36}$$

1. $$x \gt y$$
2. $$x \lt y$$
3. $$x \ge y$$
4. $$x \le y$$
5. $$x = y$$ or No relation can be established between $$x$$ and $$y$$

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

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

1. (I). $$x^2 - \sqrt{25} = \sqrt{16}$$, (II). $$y^2 - 6y + 9 = 0$$

1. $$x \gt y$$
2. $$x \lt y$$
3. $$x \ge y$$
4. $$x \le y$$
5. $$x = y$$ or No relation can be established between $$x$$ and $$y$$

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

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

1. (I). $$2x^2 - 6x + 4 = 0$$, (II). $$y^2 - 12y + 32 = 0$$

1. $$x \gt y$$
2. $$x \lt y$$
3. $$x \ge y$$
4. $$x \le y$$
5. $$x = y$$ or No relation can be established between $$x$$ and $$y$$

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

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